Mirrors/Zines
1
0
Fork 0
mirror of https://github.com/fdiskyou/Zines.git synced 2025-03-09 00:00:00 +01:00
Code Issues Releases Wiki Activity
Zines/phrack/68/8.txt

3029 lines
125 KiB
Text
Raw Permalink Normal View History

1st import into tree
2016-12-14 22:07:35 +00:00
==Phrack Inc.==
Volume 0x0e, Issue 0x44, Phile #0x08 of 0x13
|=-----------------------------------------------------------------------=|
|=--------=[ Practical cracking of white-box implementations ]=---------=|
|=-----------------------------------------------------------------------=|
|=---------------=[ by SysK - whiteb0x [o] phrack o org ]=---------------=|
|=-----------------------------------------------------------------------=|
-------
1 - Introduction
2 - What is a WB implementation?
3 - The things you should know about white-boxes
3.1 - Products available
3.2 - Academic state of the art
4 - Handling the first case: hack.lu's challenge
4.1 - The discovery step
4.2 - The key recovery
4.3 - Random thoughts
5 - White-boxing the DES
5.1 - The DES algorithm
5.2 - An overview of DES WB primitives
6 - Breaking the second case: Wyseur's challenge
6.1 - Efficient reverse engineering of the binary
6.2 - The discovery step
6.3 - Recovering the first subkey
6.4 - Recovering the original key
7 - Conclusion
8 - Gr33tz
9 - References
10 - Appendix: Source code
-------
--[ 1 - Introduction
This paper is about WB (white-box) cryptography. You may not have heard
too much about it but if you're focused on reverse engineering and more
precisely on software protections, then it may be of interest for you.
Usually The common way to learn something valuable in cryptography is
either to read academic papers or cryptography books (when they're written
by true cryptographers). However as cryptography is about maths, it can
sometimes seem too theoretical for the average reverser/hacker. I'm willing
to take a much more practical approach using a combination of both reverse
engineering and elementary maths.
Obviously such a paper is not written for cryptographers but rather for
hackers or crackers unfamiliar with the concept of white-box and willing to
learn about it. Considering the quasi non existence of public
implementations to play with as well as the 'relatively' small amount of
valuable information on this subject, I hope this will be of interest. Or
at the very least that it will be a pleasant read... O:-)
--[ 2 - What is a WB implementation?
So let's begin with a short explanation. A white-box is a particular
implementation of a cryptographic primitive designed to resist to the
examination of its internals. Consider the case of a binary embedding (and
using) a symmetric primitive (such as AES for example). With the common
implementations, the AES key will always leak in memory at some point of
the execution of the program. This is the classic case of a reverser using
a debugger. No matter how hard it may be (anti-debug tricks, obfuscation of
the key, etc.), he will always find a way to intercept the key. White-box
cryptography techniques were designed to solve this problematic which is
very common, especially in the field of DRM (Digital Rights Management).
So how does it work? The main concept that you should remember is that
the key is never explicit. Or you could say that it's mathematically
transformed or 'fused' with the encryption routine. So for one key there is
one particular obfuscated primitive which is strictly equivalent to the
original one*. For a same input, both implementations will produce an
identical output. The mathematical transformation is designed in such a way
that an attacker with a debugger will not be able to deduce the key from
the internal state ... at least in a perfect world :-)
*: It's not 'exactly' true as we will see later with external encodings.
Confused? Then take a look at this tiny example:
-> Function1: for x in [0:3] f(x) = (k+x) % 4
-> Function2: for x in [0:3] g(x) = S[x] with S = [3,0,1,2]
If k==3, then the two functions f() and g() are equivalent. However the
first one explicitly uses the key 'k' whereas the second one doesn't, being
implemented as a lookup table (LUT). You could say that g() is a white-box
implementation of f() (albeit a very weak one) for the key 3. While this
example is easy to understand, you will soon discover that things are more
complicated with the obfuscation of a whole real life crypto primitive.
--[ 3 - The things you should know about white-boxes
<<<<<<<<<<<<<<<<<< DISCLAIMER <<<<<<<<<<<<<<<<<<
> I will voluntarily not enter into too much <
> details. As I said, the paper is based on a <
> practical approach so let's avoid the maths <
> for the moment. <
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
----[ 3.1 Products available
WB cryptography is essentially implemented in commercial security
products by a relatively small number of companies (Cloakware -acquired by
Irdeto-, Whitecryption, Arxan, Syncrosoft, etc.). Usually they provide a
secure API which is then integrated into other security primitives, often
for DRM purposes. Amongst other things, they design WB primitives for
symmetric encryption (DES, AES) but also MAC (HMAC, CMAC) and asymmetric
primitives (ECC, RSA, DSA).
How often do we come across WB in the wild? More than you could think
of! For example you can see in [R10] that Irdeto has many famous customers
including TI, Sony, Adobe and NetFLIX. WB cryptography will most likely
become more and more present in software protections.
As far as I can tell, there are unfortunately only 2 public (non
commercial) examples of WB implementations, both with undisclosed
generators:
- The first historical one is a binary available on Brecht Wyseur's
website [R04] and is a WB DES implementation. Brecht challenges
people to find the key:
"If you like, try to extract the secret key, using all information
you can find from this implementation (besides brute-force black-box
attacks of course)."
Keep in mind that this is a challenge, not some production code.
- The second one less likely to be known is a challenge proposed by Jb
for the 2009 edition of hack.lu [R02]. This one is a simplistic AES
WB but was never labeled officially as such. Part of the challenge is
indeed to find out (oops!).
The cryptanalysis involved is obviously far below the academic state of
the art but it's nonetheless an interesting one and a first step for who
wants to be serious and aims at defeating more robust implementations.
We'll study both starting with Jb's binary and see how the solution can
be found in each case.
,---.
,.'-. \
( ( ,'"""""-.
`,X `.
/` ` `._
( , ,_\
| ,---.,'o `.
| / o \ )
\ ,. ( .____,
\| \ \____,' \
'`'\ \ _,____,'
\ ,-- ,-' \
( C ,' \
`--' .' |
| | .O |
__| \ ,-'_
/ `L `._ _,' ' `.
/ `--.._ `',. _\ `
`-. /\ | `. ( ,\ \
_/ `-._ / \ |--' ( \
' `-. `' \/\`. `. )
\ -hrr- \ `. | |
----[ 3.2 Academic state of the art
AFAIK academic publications are limited to symmetric encryption and
especially to DES and AES (though SPN case is somewhat extended in [R08]).
Explaining the history of the design and the cryptanalysis techniques which
were developed would be complicated and is already explained with great
details in the thesis of Brecht Wyseur [R04].
The main question is to know if there exists a secure WB design and if
you consider the current state of the art in cryptography, well... there
isn't! There is currently no implementation proposal not broken by design.
And in this case, broken means a key recovery in a matter of seconds in the
worst case. Yes, _that_ broken!
However, real-life white-box cryptography may be different because:
- As explained before, proprietary implementations of algorithms
not mentioned in any paper (MAC algorithms, asymmetric ones)
exist. This proves that people were smart enough to design new
implementations. On the other hand, without any formal analysis
of these implementations, nothing can be said regarding their
effective security.
- Cloakware products were at least partially designed/written by
the cryptographers who designed the first white-box [R7]. On one
hand you may suspect that their product is broken by design.
Alternatively it can be assumed that it is at least immune
against current cryptanalysis techniques. Little can be said
about other protections (whitecryption, Arxan, Syncrosoft) but we
could speculate that it's not of the same caliber.
So are WB protections hard to break in practice? Who knows? But
remember that protecting the key is one thing while protecting a content is
something different. So if you ever audit a white-box solution, before
trying to retrieve the key, see if you can intercept the plaintexts. There
are lots of possible attacks, potentially bypassing the WB protections
[R06].
Remark: Obviously in the case of DRM (if no hardware protection is
involved), you will always find a way to intercept unencrypted data. This
is because at some point the player will have to send audio/video streams
to the sound/video card drivers and you may want to hook some of their
functions to recover the media. This is however a practice to forget if the
media requires the synchronization of several streams (i.e. movies with
both audio and video).
Now that said, let's begin with the first challenge :)
--[ 4 - Handling the first case: hack.lu's challenge
I have to thank Jb for this binary as he was the one who suggested me
to solve it a few days ago*. Unfortunately my solution is biased as I knew
from the very beginning that it was an AES white-box. I may have taken a
different approach to solve it if I hadn't. This is however a good enough
example to introduce WB protections.
*: Phrack being usually late "a few days ago" probably means "a few weeks**
ago"
**: Phrack being _indeed_ late "a few weeks ago" is now "a few months ago"
;>
----[ 4.1 - The discovery step
Since the challenge is about breaking an AES white-box, it means that
we may need to perform several tasks:
- finding out if the WB is an AES or an AES^-1 and the associated key
length: 16 (AES-128), 24 (AES-192), 32 (AES-256)? We want to discover
exactly *what* was white-boxed.
- reversing every cryptographic functions involved and discovering how
they are related to the original AES functions. This is about
understanding *how* the implementation was white-boxed.
- finding a way to recover the original key.
I won't describe the AES as it's not necessary to understand this part.
The necessary details will be provided a bit later. First of all, let's see
how the serial is retrieved. We'll start by a quick reverse engineering of
the sub_401320() function:
---------------------------------------------------------------------------
mov eax, [esp+38h+hDlg]
push 21h ; cchMax
lea ecx, [esp+3Ch+String]
push ecx ; lpString
push 3ECh ; nIDDlgItem
push eax ; hDlg
call ds:GetDlgItemTextA
cmp eax, 20h ; is length == 32?
---------------------------------------------------------------------------
Without too much surprise, GetDlgItemText() is called to retrieve an
alpha-numeric string. The comparison in the last line implies a length of
32 bytes in its ASCII representation (not including the null byte) hence a
16 bytes serial. Let's continue:
---------------------------------------------------------------------------
cmp eax, 20h
jz short good_serial ; if len is ok then start the
; conversion
bad_serial:
xor eax, eax
[...]
retn ; return 0
good_serial:
push ebx
push esi
xor esi, esi ; i=0
nop
build_data_buffer:
movzx edx, [esp+esi*2+40h+String]
push edx
call sub_4012F0 ; get least significant nibble
mov ebx, eax
movzx eax, [esp+esi*2+44h+var_27]
push eax
shl bl, 4
call sub_4012F0 ; get most significant nibble
or bl, al ; bl is now a converted byte
mov byte ptr [esp+esi+48h+input_converted], bl
; input_converted[i] = bl
inc esi ; i++
add esp, 8
cmp esi, 10h
jl short build_data_buffer
lea ecx, [esp+40h+input_converted]
push ecx
mov edx, ecx
push edx
call sub_401250 ; white-box wrapper
add esp, 8
pop esi
mov eax, 10h
xor ecx, ecx
pop ebx
; Compare the resulting buffer byte after byte
compare_buffers:
mov edx, [esp+ecx+38h+input_converted]
cmp edx, dword ptr ds:aHack_lu2009Ctf[ecx]
; "hack.lu-2009-ctf"
jnz short bad_serial
sub eax, 4
add ecx, 4
cmp eax, 4
jnb short compare_buffers
[...]
retn
---------------------------------------------------------------------------
The alphanumeric string is then converted byte after byte using the
sub_4012F0() function in the corresponding plaintext (or ciphertext) block
for cryptographic manipulations. The function sub_401250() is then called
taking it as a parameter. When the function returns, the buffer is then
compared to the "hack.lu-2009-ctf" string (16 bytes). If both are equal,
the serial is valid (the function returns 1).
Let's see sub_401250() in more detail:
---------------------------------------------------------------------------
sub_401250 proc near ; WrapperWhiteBox
[...]
mov eax, [esp+14h+arg_0]
push esi
mov esi, [esp+18h+arg_4]
xor ecx, ecx
add eax, 2
lea esp, [esp+0]
permutation1:
; First step is a transposition (special permutation)
movzx edx, byte ptr [eax-2]
mov [esp+ecx+18h+var_14], dl
movzx edx, byte ptr [eax-1]
mov [esp+ecx+18h+var_10], dl
movzx edx, byte ptr [eax]
mov [esp+ecx+18h+var_C], dl
movzx edx, byte ptr [eax+1]
mov [esp+ecx+18h+var_8], dl
inc ecx
add eax, 4
cmp ecx, 4
jl short permutation1
; Second step is calling the white-box
lea eax, [esp+18h+var_14]
push eax
call sub_401050 ; call WhiteBox
[...]
permutation2:
; Third step is also a transposition
; Bytes' position are restored
movzx edx, [esp+ecx+14h+var_14]
mov [eax-2], dl
movzx edx, [esp+ecx+14h+var_10]
mov [eax-1], dl
movzx edx, [esp+ecx+14h+var_C]
mov [eax], dl
movzx edx, [esp+ecx+14h+var_8]
mov [eax+1], dl
inc ecx
add eax, 4
cmp ecx, 4
jl short permutation2
[...]
retn
---------------------------------------------------------------------------
At first sight, sub_401250() is composed of three elements:
- A first bunch of instructions operating on the buffer which is
no more than a (4x4) matrix transposition operating on bytes.
For example:
A B C D A E I M
E F G H becomes B F J N
I J K L C G K O
M N O P D H L P
This is a common step to prepare the plaintext/ciphertext block
into the so-called "state" as the AES is operating on 4x4 matrix.
- This function is then calling sub_401050() which is composed of
elementary operations such as XOR, rotations and substitutions.
- A second transposition. One important thing to know about the
transposition is that the function is its own inverse. The former
bytes' positions are thus restored.
sub_401050() is the WB. Whether it's an AES or an AES^-1 function and
its keylength has yet to be determined. The serial acts as a plaintext or
a ciphertext which is (de,en)crypted using a key that we want to retrieve.
Since the output buffer is compared with an English sentence, it seems fair
to assume that the function is an AES^-1.
Reverse engineering of sub_401050()
-----------------------------------
Detailing the whole reverse engineering steps is both boring and
meaningless as it doesn't require special skills. It's indeed pretty
straightforward. The resulting pseudo C code can be written as such:
----------------------------- First version -------------------------------
void sub_401050(char *arg0)
{
int round,i;
// 9 first rounds
for(round=0; round<9; round++)
{
// step-1(round)
for(i=0; i<16; i++)
arg0[i] = (char) 0x408138[ i + (arg0[i] + round*0x100)*16 ];
// step-2
sub_401020(arg0);
// step-3
for(i=0; i<4; i++)
{
char cl,dl, bl, var_1A;
cl = byte_414000[ arg0[0+i]*4 ];
cl ^= byte_414400[ arg0[4+i]*4 ];
cl ^= byte_414800[ arg0[8+i]*4 ];
cl ^= byte_414C00[ arg0[12+i]*4 ];
dl = byte_414000[ 1 + arg0[0+i]*4 ];
dl ^= byte_414400[ 1 + arg0[4+i]*4 ];
dl ^= byte_414800[ 1 + arg0[8+i]*4 ];
dl ^= byte_414C00[ 1 + arg0[12+i]*4 ];
bl = byte_414000[ 2 + arg0[0+i]*4 ];
bl ^= byte_414400[ 2 + arg0[4+i]*4 ];
bl ^= byte_414800[ 2 + arg0[8+i]*4 ];
bl ^= byte_414C00[ 2 + arg0[12+i]*4 ];
var_1A = bl;
bl = byte_414000[ 3 + arg0[0+i]*4 ];
bl ^= byte_414400[ 3 + arg0[4+i]*4 ];
bl ^= byte_414800[ 3 + arg0[8+i]*4 ];
bl ^= byte_414C00[ 3 + arg0[12+i]*4 ];
arg0[0+i] = cl;
arg0[4+i] = dl;
arg0[8+i] = var_1A;
arg0[12+i] = bl;
}
}
// step-4
for(i=0; i<16; i++)
arg0[i] = (char) 0x411138 [ i + arg0[i] * 16 ]
// step-5
sub_401020(arg0);
return;
}
----------------------------- First version -------------------------------
It seems that we have a 10 (9 + 1 special) rounds which probably means
an AES-128 or an AES-128^-1 (hence a 16 bytes keylength as both are
related).
Remark: Something very important is that we will try to solve this problem
using several assumptions or hypotheses. For example there is no evident
proof that the number of rounds is 10. It _seems_ to be 10 but until the
functions (and especially the tables) involved are not analyzed, we should
always keep in mind that we may be wrong with the guess and that some evil
trick could have been used to fool us.
Now we have the big picture, let's refine it a bit. For that, we will
analyze:
- The tables at addresses 0x408138 (step-1) and 0x411138 (step-4)
- The round independent function sub_401020 (step-2, step-5)
- step-3 and the 16 arrays byte_414x0y with:
- x in {0,4,9,C}
- y in {0,1,2,3}
The tables are quite easy to analyze. A short look at them show that
there is one substitution table per character per round. Each substitution
seems to be a "random" bijection. Additionally, 0x408138 + 16*256*9 =
0x411138 (which is the address of the last round's table).
The function sub_401020() is a mere wrapper of function sub_401000():
---------------------------------------------------------------------------
void sub_401020(arg0)
{
int i;
for(i=0; i<4; i++)
sub_401000(arg0, 4*i, i);
}
// arg4 parameter is useless but who cares?
void sub_401000(arg0, arg4, arg8)
{
if(arg8 != 0)
{
(int) tmp = ((int)arg0[4*arg8] << (8*arg8)) & 0xFFFFFFFF;
(int) arg0[4*arg8] = tmp | ((int)arg0[4*arg8] >> (32-(8*arg8)));
}
return;
}
---------------------------------------------------------------------------
This is clearly the ShiftRows() elementary function of the AES.
For example:
59 49 90 3F 59 49 90 3F [ <<< 0 ]
30 A7 02 8C becomes A7 02 8C 30 [ <<< 1 ]
0F A5 07 22 07 22 0F A5 [ <<< 2 ]
F9 A8 07 DD DD F9 A8 07 [ <<< 3 ]
here '<<<' is a cyclic shift
ShiftRows() is used in the encryption function while the decryption
function uses its inverse. Unless there is a trap to fool us, this is a
serious hint that our former assumption was wrong and that the WB is an
AES, not an AES^-1.
Now regarding step-3 let's begin by looking at the tables. They all
hold bijections but clearly neither random nor distinct ones. Let's look
for example at the byte_414400 table:
byte_414400 : 0 3 6 5 C F A 9 ...
(The elements of this table are located at 0x414400, 0x414404,
0x41440C, etc. This is because of the *4 that you can see in the C
code. This rule also applied to the 15 other tables.)
If you ever studied/implemented the AES then you must know that its
structure is algebraic. The MixColumns in particular is an operation
multiplying each columns of the state by a particular 4x4 matrix. The
coefficients of such mathematical objects are _not_ integers but rather
elements of GF(2^8) whose representation is fixed by a particular binary
polynomial.
Now if you don't have a clue about what I'm saying let's just say that
the multiplication of said AES coefficients is not a simple integer
multiplication. Since the calculus in itself would be highly inefficient
most implementations use special tables holding the precomputed results.
AES requires to know how to multiply by 01, 02, and 03 in GF(2^8). In
particular byte_414400 is a table used to compute b = 3*a in such field (a
is the index of the table and b is the value stored at this index).
Now let's look at the tables. In each case it was easy to see that they
were holding a precomputed multiplication by a given coefficient:
byte_414000 : 0 2 4 6 8 A C E ... // Coef = 2
byte_414400 : 0 3 6 5 C F A 9 ... // Coef = 3
byte_414800 : 0 1 2 3 4 5 6 7 ... // Coef = 1
byte_414C00 : 0 1 2 3 4 5 6 7 ... // Coef = 1
byte_414001 : 0 1 2 3 4 5 6 7 ... // Coef = 1
byte_414401 : 0 2 4 6 8 A C E ... // Coef = 2
byte_414801 : 0 3 6 5 C F A 9 ... // Coef = 3
byte_414C01 : 0 1 2 3 4 5 6 7 ... // Coef = 1
byte_414002 : 0 1 2 3 4 5 6 7 ... // Coef = 1
byte_414402 : 0 1 2 3 4 5 6 7 ... // Coef = 1
byte_414802 : 0 2 4 6 8 A C E ... // Coef = 2
byte_414C02 : 0 3 6 5 C F A 9 ... // Coef = 3
byte_414003 : 0 3 6 5 C F A 9 ... // Coef = 3
byte_414403 : 0 1 2 3 4 5 6 7 ... // Coef = 1
byte_414803 : 0 1 2 3 4 5 6 7 ... // Coef = 1
byte_414C03 : 0 2 4 6 8 A C E ... // Coef = 2
As a result, step-3 can be written as:
[ arg0(0,i) [ 02 03 01 01 [ arg0(0,i)
arg0(4,i) = 01 02 03 01 x arg0(4,i)
arg0(8,i) 01 01 02 03 arg0(8,i)
arg0(c,i) ] 03 01 01 02 ] arg0(c,i) ]
And this is exactly the MixColumns of the AES! Everything taken into
account gives this new version of sub_401250():
---------------------------- Second version -------------------------------
void sub_401050(char *arg0)
{
int round,i;
// 9 first rounds
for(round=0; round<9; round++)
{
// step-1(round)
for(i=0; i<16; i++)
arg0[i] = (char) 0x408138[ i + (arg0[i] + round*0x100)*16 ];
// step-2
ShiftRows(arg0);
// step-3
MixColumns(arg0);
}
// Last round
// step-4
for(i=0; i<16; i++)
arg0[i] = (char) 0x411138 [ i + arg0[i]*16 ];
// step-5
ShiftRows(arg0);
return;
}
---------------------------- Second version -------------------------------
This confirms the assumption that the WB is an AES as AES^-1 uses the
invert function of MixColumns which makes use of a different set of
coefficients (matrix inversion). As you can see the key material is not
explicit in the code, somehow hidden in the tables used in step-1. Kinda
normal for a WB ;)
----[ 4.2 - The key recovery
The general algorithm (not including the key schedule which generates K)
of AES-128 encryption is the following:
---------------------------------------------------------------------------
ROUNDS=10
def AES_128_Encrypt(in):
out = in
AddRoundKey(out, K[0])
for r in xrange(ROUNDS-1):
SubBytes(out)
ShiftRows(out)
MixColumns(out)
AddRoundKey(out,K[r])
SubBytes(out)
ShiftRows(out)
AddRoundKey(out, K[10])
return out
---------------------------------------------------------------------------
Where K[r] is the subkey (16 bytes) used in round r. From now on, 'o'
is the symbol for the composition of functions, this allows us to write:
SubBytes o AddRoundKey(K[r],IN) = step-1(IN,r) for r in [0..9]
Exploiting the first round, this immediately gives a system of
equations (with S being located at 0x408138):
SubBytes(K[0][i] ^ arg0[i]) = S[ i + arg0[i]*16 ] for i in [0..15]
The equations hold for any arg0[i] and in particular for arg0[i] = 0.
The resulting simplified system is thus:
SubByte(K[0][i]) = S[i] for i in [0..15]
K[0][i] = SubByte()^-1 o S[i] for i in [0..15]
Let's try it on the rub^Wpython console:
---------------------------------------------------------------------------
>>> sbox2 = inv_bij(sbox); # We compute SubBytes^-1
>>> S = [0xFA, 0xD8, 0x88, 0x91, 0xF1, 0x93, 0x3B, 0x39, 0xAE, 0x69, 0xFF,
0xCB, 0xAB, 0xCD, 0xCF, 0xF7]; # dumped @ 0x0408138
>>> for i in xrange(16):
... S2[i] = sbox2[S2[i]];
...
>>> S2;
[20, 45, 151, 172, 43, 34, 73, 91, 190, 228, 125, 89, 14, 128, 95, 38]
---------------------------------------------------------------------------
But remember that a transposition is necessary to retrieve the subkey!
---------------------------------------------------------------------------
>>> P = [0, 4, 8, 12, 1, 5, 9, 13, 2, 6, 10, 14, 3, 7, 11, 15] #I'm lazy :)
>>> S4 = []
>>> for i in xrange(16):
... S4.insert(i,S2[P[i]])
---------------------------------------------------------------------------
Now S4 holds the subkey K[0]. An interesting property of the AES key
schedule is that the subkey K[0] is equal to the key before derivation.
This is why our priority was to exploit the first round.
---------------------------------------------------------------------------
>>> s = 'hack.lu-2009-ctf'
>>> key = ''.join(map(lambda x: chr(x), S4))
>>> key
'\x14+\xbe\x0e-"\xe4\x80\x97I}_\xac[Y&'
>>> keyObj=AES.new(key)
>>> encPwd=keyObj.decrypt(s)
>>> encPwd.encode('hex').upper()
'192EF9E61164BD289F773E6C9101B89C'
---------------------------------------------------------------------------
And the solution is the same as baboon's one [R03]. Of course there
were many other ways to proceed but it's useless to even consider them due
to the very weak nature of this 'WB'.
----[ 4.3 - Random thoughts
Jb designed this challenge so that it could be solved in the 2-days
context of the hack.lu CTF. It's very likely that any reverser familiar
with the AES would be able to deal with it rather easily and so did baboon
at that time when he came up with a smart and quick solution [R03]. If Jb
had used the implementation described in [R07] then it would have been a
whole other game though still breakable [R05].
That being said, this implementation (which is based on what is called
partial evaluation) may only be a toy cipher but it's perfect to introduce
more advanced concepts. Indeed several security measures (amongst others)
were voluntary missing:
- ShiftRows() and MixColumns() were not modified. A strong
implementation would have transformed them. Additionally SubBytes()
could have been transformed in a less gentle manner to mitigate
trivial attacks.
- There is a direct correspondence between an obfuscated function and
it's unprotected "normal" counterpart. Inputs and outputs of such
functions are synchronized or you could say that intermediate states
can be observed. "Internal encoding" removes this property.
- The first and last rounds should have a special protection. This is
because the input (respectively the output) of the first
(respectively the last) round can be synchronized with the normal
implementation. "External encoding" is used to prevent this but as a
side effect alter the compatibility with the original encryption.
- etc.
Remark: If you ever come across a WB implementation, let me give you 2 nice
tricks to see in the blink of an eye if it's potentially weak or not:
- Look at the size of the implementation. Remember that the size of an
obfuscated primitive is deeply related to the number and size of the
lookup tables, the weight of the opcodes being generally negligible.
In this case, the binary was 85 kB whereas the state of the art
requires at least 770 kB. It was thus obvious that several
obfuscation layers were missing.
- The fully obfuscated version of the algorithms described in [R07]
only uses XOR and substitutions (lookup tables) as MixColumns and
ShiftRows are both transformed to make it possible. One may however
point out that the statement holds with T-tables based
implementations. It's true but such implementations use well known
tables so it's easy to fingerprint them.
Remember that real-life white-boxes (i.e. used in DRM, embedded
devices, etc.) are likely to be close to the state of the art (assuming
they are designed by crypto professionals and not by the average engineer
;)). Conversely, they may also face practical problematics (size, speed)
which have an impact on their security. This is especially true with
embedded devices.
--[ 5 - White-boxing the DES
If you're still reading (hi there!) then it probably means that you
already have at least a basic knowledge of cryptography. So you know that
DES should not be used because of its short keylength (56 bits), right?
Then why the hell should we be focused on it? Well because:
- There are only 2 public white-box design families: AES and DES
- If you can white-box DES, then you can probably white-box 3DES as
well (which is strong)
- I couldn't find a non commercial sophisticated enough AES WB to play
with and I don't want to be sued by M$, Adobe, etc. :D
Remark: While AES WB cryptanalysis are essentially algebraic, DES related
ones are statistical as you will soon find out.
----[ 5.1 - The DES algorithm
DES is a so called Feistel cipher [R01] with a block size of 64 bits
and 16 rounds (r). First a permutation (IP) is applied to the input then
in each round a round-function is applied which splits its input in two 32
bits buffers L (Left) and R (Right) and applies the following equations
system:
L(r+1) = R(r)
R(r+1) = L(r) [+] f(R(r),K(r))
With:
0 <= r < 16
[+] being the XOR operation
The round function is described by the following scheme:
--------------------------- DES round function ----------------------------
********** **********
* L(r) * * R(r) *
********** **********
| |
.------------- |
| | v .---------------------.
| | .-------------. / Linear transformation \
| | \ E-Box / ( 32b -> 48b )
| | '---------' \ /
| | | '---------------------'
| | v .------------.
| | ..... ********** / XOR operand \
| | . + .<------ * K(r) * ( 2x48b -> 48b )
| | ..... ********** \ /
| | /\ '------------'
| | / \
| | v v .-------------------------.
| | .------. .------. / Non linear transformation \
| | \ S1 / ... \ S8 / ( using SBox )
| | '--' '--' \ 8x6b -> 8x4b /
| | \ / '-------------------------'
| | \ /
| | v v .---------------------.
| | .--------. / Linear transformation \
| | | P-Box | ( 32b -> 32b )
| | '--------' \ /
| | | '---------------------'
| | ..v.. .------------.
| '--------->. + . / XOR operand \
| ..... ( 2x32b -> 32b )
| | \ /
v v '------------'
********** **********
* L(r+1) * * R(r+1) *
********** **********
---------------------------------------------------------------------------
When the 16 rounds are completed, the IP^-1() function is applied and
the result is called the ciphertext.
While SBox and XOR are self explanatory, let me give you a few more
details about the linear transformations (E-Box and P-Box).
The E-Box
---------
The E-Box is used to extend a 32 bits state into a 48b one so that each
bit can be combined with a round-key bit using a XOR. To transform 32 bits
into 48 bits, 16 out of the 32 bits are duplicated. This is performed using
the following table:
32, 1, 2, 3, 4, 5,
4, 5, 6, 7, 8, 9,
8, 9, 10, 11, 12, 13,
12, 13, 14, 15, 16, 17,
16, 17, 18, 19, 20, 21,
20, 21, 22, 23, 24, 25,
24, 25, 26, 27, 28, 29,
28, 29, 30, 31, 32, 1
For example, the first bit of output will the last bit of input (32)
and the second bit of output will be the first bit of input (1). In this
particular case the bit positions are written from 1 to 32. As you may have
noticed, the 16 bits from columns 3 and 4 are not duplicated. They are
called the middle bits, we will see later why they are important.
The P-Box
---------
The P-Box is a bit permutation which means that every bit of input will
have a new position in the output. Such a transformation is linear and can
be represented by a bit matrix. When combined with a XOR operation with a
constant, this is what we call an affine transformation (AT).
----[ 5.2 - An overview of DES WB primitives
The first WB DES implementation was presented in [R09]. Explaining how
and why DES white-box were designed is not the most simple of the task
especially with an ASCII / 75 columns constraint ;> I'll try to focus on
the main mechanisms so that you can get a global picture with the next
section. At some point you may however feel lost. In that case, please read
the excellent [R15] <3
The protection of I/O
---------------------
The reverser is able to intercept every step of the algorithm as well
as to examine all the memory. This gives him a huge advantage as he can
easily trace all inputs and outputs of elementary functions of the WB.
In the case of DES, this is even easier thanks to the very nature of
Feistel network. For example an attacker would easily locate the output of
the P-Box in round (r) because it is combined with part of the input: L(r).
To mitigate this, several transformations are performed:
a) All elementary operations of the white-box are performed on 96 bits
states. Let's try to understand why.
Initially a native DES round begins which the 64 bits state
L(r) || R(r). R(r) is then extended using the E-box to
generate a 8x6 = 48 bits buffer and at the same time, L(r) and R(r)
are still part of the internal state because they are still
contributing to the round's operations:
************** **************
* L(r) * * R(r) *
************** **************
| .------------| 32b
| | v
| | .-------------------.
32b | 32b | | E-box |
| | '-------------------'
| | | 48b
v v v
Mem1 Mem2 Mem3
At this point the internal state is composed of 32 x 2 + 48 = 112
bits which is the maximum required size before being shrunken to a
64 bits state at the end of the round: L(r+1) || R(r+1). To avoid
any information leak, a unique -constant size- state is used to hide
the position of the bits.
If you remember 5.1 paragraph, the E-Box is duplicating 16 out of
the 32 bits of R(r). As a result, constructing 'Mem2' can be done
extracting 16 bits out of R(r) and the 16 remaining ones out of
'Mem3'. With this property, the internal state is composed of 96
bits. Here is a diagram ripped from [R17] to understand how the
primitive is modified to handle this 96 bits state:
32b 48b 16b
************** ********************* ********
state 1: * L(r) * * X(r) * * r(r) *
************** ********************* ********
| | | |
| v | |
| ********* ..... | v
| * sK(r) *--> . + . | .-------.
| ********* ..... '-->( Merge )
| | '-------'
| v |
| .-------------. |
| \ S / |
| '---------' |
| | |
32b v v 32b 32b v
************** *************** ***************
state 2: * L(r) * * Y(r+1) * * R(r) *
************** *************** ***************
| | |
v | |
..... .--------. | |
. + .<---| P |<-' |
..... '--------' |
| |
32b '----------------------------------.
| | |
.-------------------|-----------' |
| 32b v v 32b
| .-------. .------.
| / E-box \ ( Select )
| 32b '-----------' '------'
| | |
v 48b v v 16b
************** ********************* ********
state 3: * L(r+1) * * X(r+1) * *r(r+1)*
************** ********************* ********
With:
- sK(r) being the subkey of round r
- X(r) being the output of the E-box of round r-1
- Y(r) being the output of the SBox of round r-1
- r(r) being the complementary bits so that X(r) and r(r) is a
duplication of R(r)
b) Input and outputs between elementary operations are protected using
what is called the "internal encodings". These encodings are applied
to functions implemented as lookup tables.
Let's take an example. You are chaining f() and g() which means that
you are calculating the composition g() o f(). Obviously without any
protection, an attacker can intercept the result of f() at debug
time (e.g. by putting a breakpoint at the entry of g())
Now if you want to protect it, you can generate a random bijection
h() and replace f() and g() by F() and G() where:
F() = h() o f()
G() = g() o h()^-1
Note: Again this is a mere example, we do not care about the
{co}domain consideration.
These functions are evaluated and then expressed as lookup tables.
Obviously this will not change the output as:
G() o F() = (g() o h()^-1) o (h() o f())
= g() o (h()^-1 o h()) o f() [associativity]
= g() o f()
But the difference is that intercepting the output of F() doesn't
give the output of f(). Pretty cool trick, right?
However I've just written that WB DES implementations were always
manipulating 96 bits states. Then does it mean that we need lookup
tables of 2^96 entries? No, this would be troublesome ;> We can use
the so called "path splitting" technique.
Consider the example of a 32 bits buffer. To avoid using a huge
lookup table, you can consider that this buffer is an 8 bits array.
Each element of the array will then be obfuscated using a
corresponding 8 bits lookup table as described below:
*****************************************
* IN[0] || IN[1] || IN[2] || IN[3] *
*****************************************
| | | |
| | | |
v v v v
.-------. .-------. .-------. .-------.
| 2^8 B | | 2^8 B | | 2^8 B | | 2^8 B |
'-------' '-------' '-------' '-------'
| | | |
| | | |
v v v v
*****************************************
* OUT[0] || OUT[1] || OUT[2] || OUT[3] *
*****************************************
I took the example of an 8 bits array but I could have used any
size. Something really important to understand: the smaller the
lookup table is, the more it will leak us information. Keep it in
mind.
c) Do you remember when I said that a WB implementation was the exact
implementation of the corresponding crypto primitive? Well it's not
true. Or you could say that I was simplifying things ^_~
Most of the time (in real life), WB_DES() is a G() o DES() o F()
where F() and G() are encoding functions. So the first input
(plaintext) and the last output (ciphertext) may be obfuscated as
well. This is called an "external encoding" and this is used to
harden the white-box implementation. Indeed if there were no such
functions, first & last rounds would be weaker than other rounds.
This 'academic' protection aims at preventing trivial key recovery
attacks. A WB implementation without external encoding is said to be
'naked'.
In the context of real life protections, it may (or may not) be
associated with an additional layer to protect the I/O before &
after the encryption. It would be bad to intercept the plaintext
once decrypted, right? Commercial protections almost never use
native implementations for (at least) this reason. Intercepting a
plaintext is indeed far easier than recovering the encryption key.
In the WB DES case, common academic designs use affine functions,
encoded or not.
Transforming DES functions
--------------------------
Now that we've had an overview of how I/O were protected between
elementary functions, let's see how we can build said functions.
a) The partial evaluation
This is probably the most intuitive part of the WB implementation. This
is about 'fusing' the S-Boxes with the round-keys. This is exactly what was
performed in the previous AES challenge. If you can remember, this is also
the first example that I gave at the beginning of the paper to introduce
the white-boxing concept.
Using the previous diagram, it means that we want to convert this step:
32b 48b 16b
************** ********************* ********
* L(r) * * X(r) * * r(r) *
************** ********************* ********
| | | |
| v | |
| ****** ..... | v
| * sK *--> . + . | .-------.
| ****** ..... '-->( Merge )
| | '-------'
| v |
| .-------------. |
| \ S / |
| '---------' |
| | |
32b v v 32b 32b v
************** *************** **************
* L(r) * * Y(r+1) * * R(r) *
************** *************** **************
into this one:
*********************************************
* state 1 (12 x 8 = 96 bits) *
*********************************************
| | | |
v v v v
.-----..-----..-----. .-----.
| T0 || T1 || T2 | ... | T11 |
'-----''-----''-----' '-----'
| | | |
v v v v
*********************************************
* state 2 (96 bits) *
*********************************************
A lookup table Ti (mapping a byte to a byte) is called a 'T-Box'. There
are two types of T-Box because of the heterogeneity of the operations
performed on the state:
- The non neutral T-box. They are the 8 T-boxes involved with the
Sbox and the XOR. Each of them is concealing an Sbox and a subkey
mixing.
input:
-> 6 bits from X(r) to be mixed with the subkey
-> 1 bit from L(r) or r(r)
-> 1 bit from L(r) or r(r)
output:
-> 4 bits from the Sbox
-> 2 bits from X(r) taken from the input before being
mixed with the subkey
-> 1 bit from L(r) or r(r)
-> 1 bit from L(r) or r(r)
- The neutral T-box which are only used to connect bits of state 1
to bits of state 2. For example the bits of L(r) are never
involved in any operation between state 1 and state 2.
input:
-> 1 bit from L(r) or r(r)
-> 1 bit from L(r) or r(r)
[...]
-> 1 bit from L(r) or r(r)
output:
-> the input (permuted)
Keep in mind that in each case, you have a 'nibble view' of both inputs
and outputs. Moreover, permutations are used to make harder the
localization of Sbox upon a simple observation. To have a better
understanding of this point as well as associated security explanations
I recommend to read [R09].
b) The AT transformation
We now want to transform this:
************** *************** ***************
state 2: * L(r) * * Y(r+1) * * R(r) *
************** *************** ***************
| | |
v | |
..... .--------. | |
. + .<---| P |<-' |
..... '--------' |
| |
32b '----------------------------------.
| | |
.-------------------|-----------' |
| 32b v v 32b
| .-------. .------.
| / E-box \ ( Select )
| 32b '-----------' '------'
| | |
v 48b v v 16b
************** ********************* ********
state 3: * L(r+1) * * X(r+1) * *r(r+1)*
************** ********************* ********
into this:
*********************************************
* state 2 (96 bits) *
*********************************************
| | | |
v v v ... v
?????????????????????????????????????????????
| | | ... |
v v v v
*********************************************
* state 3 (96 bits) *
*********************************************
To put it simply, and as said earlier, the combination of the P-Box and
the following XOR is an affine function. Because we want to use lookup
tables to implement it we will have to use a matrix decomposition.
Let's take an example. You want to protect a 8x8 bit-matrix
multiplication. This matrix (M) can be divided into 16 2x2 submatrix as
shown below:
.----. .----.----.----.----. .----.
| Y0 | | A | B | C | D | | X0 |
.----. .----.----.----.----. '----'
| Y1 | | E | F | G | H | | X1 |
.----. = .----.----.----.----. x .----.
| Y2 | | I | J | K | L | | X2 |
.----. .----.----.----.----. .----.
| Y3 | | M | N | O | P | | X3 |
'----' '----'----'----'----' '----'
Vector Y Matrix M Vector X
Here the Yi and Xi are 2 bits sub-vectors while A,B,C,etc. are 2x2
bit-submatrix. Let's focus on Y0, you can write:
Y0 = A*X0 [+] B*X1 [+] C*X2 [+] D*X3
Because A,B,C and D are constants it's possible to evaluate the
multiplications and build the corresponding lookup tables (Li). This gives
the following diagram:
****** ****** ****** ******
* X0 * * X1 * * X2 * * X3 *
****** ****** ****** ******
| | | |
v v v v
.----. .----. .----. .----.
| L0 | | L1 | | L3 | | L4 |
'----' '----' '----' '----'
| | | |
| ..... | | ..... |
'->. + .<-' '->. + .<-'
..... .....
| |
| ..... |
'------>. + .<------'
.....
|
v
******
* Y0 *
******
You may object (and you would be right) that information is still
leaking and that it would be easy to retrieve the original matrix. Well
it's true. Thus to avoid this kind of situation two techniques are used:
- Each XOR operation is hidden inside a lookup table. In our example,
the resulting lookup tables have 2^(2x2) = 16 entries and 2^2 = 4
outputs (hence a size of 4x16 = 64 bits).
- Internal encoding (remember the previous explanations) is used to
protect the I/O between the lookup tables.
Our matrix multiplication becomes:
****** ****** ****** ******
* X0 * * X1 * * X2 * * X3 *
****** ****** ****** ******
| | | |
v v v v
2b 2b 2b 2b
<----><----> <----><---->
.----------. .----------.
\ S0 / \ S1 /
'------' '------'
<----> <---->
2b 2b
\ /
\ /
| |
v v
2b 2b
<----><---->
.---------.
\ S2 /
'------'
<---->
2b
|
v
******
* Y0 *
******
This is called an 'encoded network'. The main side effect of this
construction is the important number of lookup tables required.
--[ 6 - Breaking the second case: Wyseur's challenge
----[ 6.1 - Reverse engineering of the binary
As far as I can tell, there is an obvious need to rewrite the binary as
C code because:
- We need to understand exactly what's going on from a mathematical
point of view and C is more suitable than ASM for that purpose
- Rewriting the functions will allow us to manipulate them easily
with our tools. This is not mandatory though because we could
be using debugging functions on the original binary itself.
Again I won't detail all the reverse engineering process because this
is neither the main topic nor very hard anyway compared to what you may
find in the wild (in commercial protections).
High level overview
--------------------
Let's begin by running the executable:
---------------------------------------------------------------------------
$ ./wbDES.orig
Usage: ./wbDES.orig <INPUT>
Where <INPUT> is an 8-byte hexadecimal representation of the input to be
encrypted
Example: ./wbDES.orig 12 32 e7 d3 0f f1 29 b3
---------------------------------------------------------------------------
OK so we need to provide the 8 bytes of the plaintext as separate
arguments in the command line. Hum, weird but OK. When the binary is
executed, the first thing performed is a conversion of the arguments
because obviously a suitable buffer for cryptographic operations is
necessary. The corresponding instructions were rewritten as the following
C function:
---------------------------------------------------------------------------
// I even emulated a bug, will you find it? ;>
inline void convert_args(char **argv)
{
int i = 0; // ebp-50h
while(i <= 7)
{
char c;
c = argv[1+i][0];
if(c <= '9')
{
c -= '0'; // 0x30 = integer offset in ASCII table
in[i] = (c<<4);
}
else
{
c -= ('a' - 10);
in[i] = (c<<4);
}
c = argv[1+i][1];
if(c <= '9')
{
c -= '0'; // 0x30 = integer offset in ASCII table
in[i] ^= c;
}
else
{
c -= ('a' - 10);
in[i] ^= c;
}
i++;
}
return;
}
---------------------------------------------------------------------------
Once the job is done, an 8 bytes buffer (in[8], the plaintext) is
built. This is where serious business begins. Thanks to the Control Flow
Graph provided by your favorite disassembler, you will quickly identify 3
pseudo functions* :
- wb_init(): 0x0804863F to 0x08048C1D
This code takes an 8 bytes buffer, returns 1 byte and is called 12
times by main(). Thanks to this, a 12 x 8 = 96 bits buffer is built.
As said earlier, the WB is operating on 96 bits states so this is
most likely the initialization function.
- wb_round(): 0x08048C65 to 0x08049731
This code takes the 12 bytes buffer generated by wb_init() as input
and modifies it. The function is called 16 times by main(). Because
16 is exactly the number of DES rounds, assuming it is the round
function seems fair.
- wb_final(): 0x08049765 to 0x08049E67
This code takes the last buffer returned by wb_round() as input and
returns an 8 bytes buffer which is printed on the screen. So we can
assume that this is the termination function in charge of building
the DES ciphertext out of the last internal state.
*: There is no 'function' in this program, probably because of an inlining,
but we can still distinguish logical parts.
You may argue that attributing roles to wb_init, wb_round and wb_final
is a bit hasty but there is something interesting in the code: symbols! In
each of these functions, an array of lookup tables is used and named
'Initialize', 'RoundAffineNetwork' and 'FinalRoundNetwork' in the
corresponding functions. Convenient isn't it?
Usually in commercial protections, engineers will take care of little
details such as this and try to avoid leaking any information. In this case
however, it can be assumed that the focus is on the cryptography as there
are neither anti-debugs nor anti-disassembling protections so it should be
safe to trust our intuition.
Thanks to this first reverse engineering step, we're able to rewrite
a similar main function:
-------------------------------- wb_main.c --------------------------------
unsigned char in[8]; // ebp-1Ch
unsigned char out[12]; // ebp-28h
unsigned char temp[12]; // ebp-34h
[...]
int main(int argc, char **argv)
{
if( argc != 9)
{
printf(usage, argv[0], argv[0]);
return 0;
}
/* Fill the in buffer */
convert_args(argv);
/* and print it :) */
printf("\nINPUT: ");
for(j=0; j<8; j++)
printf("%02x ", in[j]);
/* WB initialisation */
for(j=0; j<12; j++)
wb_init(j);
round_nbr = 0;
for(round_nbr=0; round_nbr<15; round_nbr++)
{
memcpy(temp, out, 12);
wb_round();
}
/* Building the final buffer */
printf("\nOUTPUT: ");
for(j=0; j<8; j++)
wb_final(j);
printf("\n");
return 0;
}
-------------------------------- wb_main.c --------------------------------
One hint to speed up things: always focus first on the size and nature
of buffers transmitted to the different sub-functions.
Reversing wb_init()
-------------------
It is now time to have a look at the first function. Again I won't
detail the whole reverse engineering but rather give you a few
explanations:
- Whenever the function is called, it uses a set of 15 lookup tables
whose addresses are dependent of both the index in the output array
and the index of the box itself (amongst the 15 used by the
function).
This means that the sets of tables used to calculate OUT[x] and
OUT[y] when x!=y are (likely to be) different and for a same OUT[x],
different tables will be applied to IN[a] and IN[b] if a!=b.
- All of these lookup tables are located at:
Initialize + 256*idx_box + OUT_idx*0xF00
where:
> idx_box is the index of the box amongst the 15
> OUT_idx is the index in the output array (OUT)
- The tables are static. Thanks to this property we can dump them
whenever we want. I chose to write a little GDB script (available in
appendix) to perform this task. The export is an array of lookup
tables (iBOX_i) written in C language.
- wb_init() is performing operations on nibbles (4 bits) so for a
particular output byte (OUT[m]), the generation of the 4 least
significant bits is independent of the generation of the 4 most
significant ones.
Now with this information in mind, let's have a look at the reversed
wb_init() function:
-------------------------------- wb_init.c --------------------------------
unsigned char p[8];
inline void wb_init(
int m // ebp-48h
)
{
unsigned int temp0; // ebp-228h
unsigned int temp1; // ebp-224h
[...]
unsigned int temp23; // ebp-1CCh
unsigned int eax,ebx,ecx,edx,edi,esi;
bzero(p,sizeof p);
p[0] = iBOX_0[m][in[0]];
p[1] = iBOX_1[m][in[1]];
p[2] = iBOX_2[m][in[2]];
p[3] = iBOX_3[m][in[3]];
p[4] = iBOX_4[m][in[4]];
p[5] = iBOX_5[m][in[5]];
p[6] = iBOX_6[m][in[6]];
p[7] = iBOX_7[m][in[7]];
// First nibble
ecx = (0xF0 & p[0]) ^ ( p[1] >> 4 );
temp3 = 0xF0 & iBOX_8[m][ecx];
ecx = (0xF0 & p[2]) ^ ( p[3] >> 4 );
eax = iBOX_9[m][ecx] >> 4;
ecx = temp3 ^ eax;
temp6 = 0xF0 & iBOX_12[m][ecx];
ecx = (0xF0 & p[4]) ^ ( p[5] >> 4 );
eax = iBOX_10[m][ecx] >> 4;
ecx = temp6 ^ eax;
edi = 0xF0 & iBOX_13[m][ecx];
ecx = (0xF0 & p[6]) ^ ( p[7] >> 4 );
eax = iBOX_11[m][ecx] >> 4;
ecx = edi ^ eax;
edx = iBOX_14[m][ecx];
esi = edx & 0xFFFFFFF0;
// Second nibble
ecx = (0x0F & p[1]) ^ (0xF0 & ( p[0] << 4 ));
temp15 = 0xF0 & ( iBOX_8[m][ecx] << 4);
ecx = (0x0F & p[3]) ^ (0xF0 & ( p[2] << 4 ));
eax = 0x0F & ( iBOX_9[m][ecx] );
ecx = temp15 ^ eax;
temp18 = 0xF0 & ( iBOX_12[m][ecx] << 4 );
ecx = (0x0F & p[5]) ^ (0xF0 & ( p[4] << 4 ));
eax = 0x0F & iBOX_10[m][ecx];
ecx = temp18 ^ eax;
temp21 = 0xF0 & (iBOX_13[m][ecx] << 4);
ecx = (0x0F & p[7]) ^ (0xF0 & ( p[6] << 4 ));
eax = 0x0F & ( iBOX_11[m][ecx] );
ecx = temp21 ^ eax;
eax = 0x0F & ( iBOX_14[m][ecx] );
// Output is the combination of both nibbles
eax = eax ^ esi;
out[m] = (char)eax;
return;
}
-------------------------------- wb_init.c --------------------------------
In a nutshell:
- & (AND) and >>/<< (SHIFTS) are used to operate on nibbles
- ^ (XOR) are used to concatenate nibbles in order to build the
entries (which are bytes) of the iBOX_i lookup tables
- The output byte out[m] is the concatenation of two independently
calculated nibbles
To understand exactly what's going on, a drawing is much clearer. So
thanks to asciio [R11] this gives us something like this:
******** ******** ******** ******** ******** ******** ******** ********
* IN_0 * * IN_1 * * IN_2 * * IN_3 * * IN_4 * * IN_5 * * IN_6 * * IN_7 *
******** ******** ******** ******** ******** ******** ******** ********
| | | | | | | |
H | H | H | H | H | H | H | H |
v v v v v v v v
<----------------------------- 8x8 = 64 bits --------------------------->
.-------..-------. .-------..-------. .-------..-------. .-------..-------.
\iBox_0 /\iBox_1 / \iBox_2 /\iBox_3 / \iBox_4 /\iBox_5 / \iBox_6 /\iBox_7 /
'-----' '-----' '-----' '-----' '-----' '-----' '-----' '-----'
<----------------------------- 8x4 = 32 bits ------------------------->
\ / \ / \ / \ /
H \ / H H \ / H H \ / H H \ / H
v v v v v v v v
.---------. .---------. .---------. .---------.
\ iBox_8 / \ iBox_9 / \ iBox_10 / \ iBox_11 /
'-------' '-------' '-------' '-------'
<------------------------- 4x4 = 16 bits ---------------------->
\ / \ /
H \ / H H \ / H
\ / \ /
v v v v
.---------. .---------.
\ iBox_12 / \ iBox_13 /
'-------' '-------'
<--------------- 2x4 = 8 bits ----------->
\ /
\ H H /
'---------. .---------'
| |
v v
.---------.
\ iBox_14 /
'-------'
<- 1x4 bits ->
\
H \ 8 bits
\ <--------->
Concatenation '---> ***********
of nibbles * OUT_x *
.---> ***********
/
L /
/
<- 1x4 bits ->
.-------.
/ iBox_14 \
'---------'
^ ^
L | | L
.--------' '--------.
/ \
/ \
<--------------- 2x4 = 8 bits ----------->
.-------. .-------.
/ iBox_12 \ / iBox_13 \
'---------' '---------'
^ ^ ^ ^
/ \ / \
L / \ L L / \ L
/ \ / \
<------------------------- 4x4 = 16 bits ---------------------->
.-------. .-------. .-------. .-------.
/ iBox_8 \ / iBox_9 \ / iBox_10 \ / iBox_11 \
'---------' '---------' '---------' '---------'
^ ^ ^ ^ ^ ^ ^ ^
L / \ L L / \ L L / \ L L / \ L
/ \ / \ / \ / \
<----------------------------- 8x4 = 32 bits ------------------------->
.-----. .-----. .-----. .-----. .-----. .-----. .-----. .-----.
/iBox_0 \/iBox_1 \ /iBox_2 \/iBox_3 \ /iBox_4 \/iBox_5 \ /iBox_6 \/iBox_7 \
'-------''-------' '-------''-------' '-------''-------' '-------''-------'
<----------------------------- 8x8 = 64 bits --------------------------->
^ ^ ^ ^ ^ ^ ^ ^
L | L | L | L | L | L | L | L |
| | | | | | | |
******** ******** ******** ******** ******** ******** ******** ********
* IN_0 * * IN_1 * * IN_2 * * IN_3 * * IN_4 * * IN_5 * * IN_6 * * IN_7 *
******** ******** ******** ******** ******** ******** ******** ********
In this case, 'H' is used as a suffix to identify the most significant
(High) nibble of a particular byte. As you can see, the input (respectively
the output) is not an 8 (respectively 12) _bytes_ array but rather a 16
(respectively 24) _nibbles_ array. Indeed, each byte array (iBox_i) stores
exactly 2 lookup tables. We say that such lookup tables are 'compacted',
see [R14] for additional details.
Global description
-------------------
Good news guys, the wb_init(), wb_round() and wb_final() functions are
composed of the same nibble oriented patterns. So basically wb_round() and
wb_final() contain also AT applied to a nibbles array and the end of the
reverse engineering is quite straightforward.
Remark: Manipulating nibbles implies that the internal encoding is
performed using 4 bits to 4 bits bijections.
Again thanks to asciio, we're able to draw something like that:
8 x (2x4) = 64 bits
<---------------------------------->
2x4 = 8 bits
<---->
.----------------------------------. .-----------.
| .-----. .-----. .-----. | | INPUT |
.----| IN0 | | IN1 | ... | IN7 | | '-----------'
| | '-----' '-----' '-----' | |
v '------------|----------------|----' v
| v | .------------.
|--------<---------------<-------' ( wb_init func )
| '------------'
.-----v---------------------------------------------. |
|.--------. .--------. .---------.| |
|| STG0_0 | | STG0_1 | ... | STG0_11 || |
|'--------' '--------' '---------'| |
'-----|---------|-----------------------------|-----' |
| | | v
| v | .-------------.
| | | ( wb_round func )
'--->-----|-------<---------------------' '-------------'
| |
.---------------|------------------------------------. |
|.--------. .---v----. .---------.| |
|| STG1_0 | | STG1_1 | ... | STG1_11 || |
|'--------' '--------' '---------'| |
'----------------------------------------------------' |
|
2x4bits |
<--------> 12 x (2x4) = 96 bits |
<----------------------------------------------------> |
v
.-------------.
... 15x ( wb_round func )
'-------------'
.----------------------------------------------------. |
|.---------..---------. .----------.| |
|| STG14_0 || STG14_1 | ... | STG14_11 || |
|'---------''---------' '----------'| |
'-----|--------|-------------------------------|-----' v
| v | .-------------.
| | | ( wb_final func )
'----->-----<----------------------------' '-------------'
| |
.-----|----------------------------. v
|.----v-. .------. .------.| .----------.
|| OUT0 | | OUT1 | ... | OUT7 || | OUTPUT |
|'------' '------' '------'| '----------'
'----------------------------------'
2x4bits
<------>
8 x (2x4) = 64 bits
<---------------------------------->
Writing the C code corresponding to these functions is not difficult
though a bit boring (not to mention prone to mistakes). I was able to
rewrite the whole binary in a few hours (and it took me almost the same
time to make it work :O). The source code is available in the appendix.
Remark: I've not tried to use Hex-Rays on the binary but it could be
interesting to know if the decompilation is working out of the box.
It's easy to see that my source code is functionally equivalent on the
input/output behavior:
---------------------------------------------------------------------------
$ ./wbDES.orig 11 22 ff dd 00 11 26 93 <- the original
INPUT: 11 22 ff dd 00 11 26 93
OUTPUT: 04 e9 ff 8e 2e 98 6c 6b
$ make
$ ./wbdes.try 11 22 ff dd 00 11 26 93 <- my copy :)
INPUT: 11 22 ff dd 00 11 26 93
OUTPUT: 04 e9 ff 8e 2e 98 6c 6b
$
---------------------------------------------------------------------------
Now let's try to break the white-box. We will proceed in two steps
which is exactly how I handled the challenge. What is described is how I
proceeded as I wasn't following academic publications. I don't know if it's
a better approach or not. It's just my way of doing things and because I'm
not a cryptographer, it's _practical_. If you prefer more _theoretical_
solutions, please refer to [R04] for a list of papers dealing with the
subject.
----[ 6.2 - The discovery step
First of all, let's gather some information about this white-box. There
is a first immediate observation: there is no explicit T-box step which
proves that it is combined with the AT step in a same function. This is an
optimization which was historically proposed in [R14] in order to protect
the output of the T-box and, as a result, to mitigate the so-called
statistical bucketing attack described in [R09] while compressing the
implementation by merging operations.
I used this information as well as the size of the binary (which is a
bit more than the size of the lookup tables) as indicators of how recent
the design could be. I didn't have the time to read all the white-box
related papers (although there are not a thousand of them).
Analyzing the wb_init()
-----------------------
Earlier, I've made assumptions about wb_init() and wb_round() but at
this point little is really known about them. Now is the time to play a bit
with wb_init() and by playing I mean discovering the "link" between the
input (plaintext) and the input of wb_round() which will be called "stage0"
from now on.
Let's begin by a quick observation. As said before, for each output
byte of wb_init(), there is a corresponding set of 14 (condensed) iBox_i.
A simple glance at these boxes is enough to determine that for each set,
the 8 first iBox_i have a very low entropy. Conversely, the remaining 5
ones have a high entropy:
---------------------------------------------------------------------------
[...]
unsigned char iBOX_3[12][256] = {
{
0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,
0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,
0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,
0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,
0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,0xf7,
[...]
0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,
0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,0xf1,
},
[...]
unsigned char iBOX_8[12][256] = {
{
0x13,0xdf,0xf9,0x38,0x61,0xe2,0x44,0x9e,0xc0,0x2a,0x0b,0xb7,0x7c,0xad,
0x56,0x85,0x96,0xbe,0x8b,0x04,0x27,0xcd,0xa8,0x1f,0xec,0x65,0x39,0xd1,
0x50,0x42,0x73,0xfa,0x4a,0x52,0x04,0x8b,0xcc,0x2f,0x19,0xad,0x67,0xe3,
[...]
0x8a,0x08,0xbd,0x59,0x36,0xf1,0xef,0x45,0x13,0xd4,0x90,0x67,0xae,0x76,
0x3c,0xf7,0xe4,0x65,0x91,0x43,0x2b,0xcd,0x80,0x58,0xd9,0x1a,0xbf,0x02,
},
[...]
---------------------------------------------------------------------------
The example shows us that iBOX_3[0] has only 2 possibles values: 0xf7
for any index inferior or equal to 127 and 0xf1 for the remaining ones.
Said otherwise, this box is a bit filter:
- High output nibble: only 1 possible value (0xf) => no bit chosen
- Low output nibble: 2 possible values (0x1, 0x7) => the input's
MSB is chosen
Let's visualize the effect of the 8 first iBox_i for every output
nibble. To see if the particular bit at position 'i' is involved in the LUT
'p' then you can compute:
- p[0]&0xf0 and p[(1<<i)]&0xf0 ; influence on the High nibble
- p[0]&0x0f and p[(1<<i)]&0x0f ; influence on the Low nibble
In each case, if the bit at the position 'i' is indeed involved then
both results will be different. I implemented it in entropy.c
(see appendix):
---------------------------------------------------------------------------
$ ./entropy
[+] Link between IN and OUT arrays
OUT[0] (high) is composed of:
-> bit 6
-> bit 49
-> bit 57
-> bit 56
OUT[0] (low) is composed of:
-> bit 24
-> bit 32
-> bit 40
-> bit 48
[...]
OUT[11] (high) is composed of:
-> bit 7
-> bit 15
-> bit 23
-> bit 31
OUT[11] (low) is composed of:
-> bit 14
-> bit 22
-> bit 46
-> bit 54
[+] Total nbr of bits involved = 96
[...]
---------------------------------------------------------------------------
So the analysis of the 8 first LUT reveals that each output (OUT[i])
nibble is linked to exactly 4 input bits. So the 8 first iBox_i are no more
than an obfuscated linear mapping.
A good idea is to focus more specifically on the input bits frequency:
---------------------------------------------------------------------------
$ ./entropy
[...]
[+] Nbr of times a bit is used
[b_00] 2 [b_01] 1 [b_02] 2 [b_03] 1 [b_04] 2 [b_05] 1 [b_06] 2 [b_07] 1
[b_08] 2 [b_09] 1 [b_10] 2 [b_11] 1 [b_12] 2 [b_13] 1 [b_14] 2 [b_15] 1
[b_16] 2 [b_17] 1 [b_18] 2 [b_19] 1 [b_20] 2 [b_21] 1 [b_22] 2 [b_23] 1
[b_24] 2 [b_25] 1 [b_26] 2 [b_27] 1 [b_28] 2 [b_29] 1 [b_30] 2 [b_31] 1
[b_32] 2 [b_33] 1 [b_34] 2 [b_35] 1 [b_36] 2 [b_37] 1 [b_38] 2 [b_39] 1
[b_40] 2 [b_41] 1 [b_42] 2 [b_43] 1 [b_44] 2 [b_45] 1 [b_46] 2 [b_47] 1
[b_48] 2 [b_49] 1 [b_50] 2 [b_51] 1 [b_52] 2 [b_53] 1 [b_54] 2 [b_55] 1
[b_56] 2 [b_57] 1 [b_58] 2 [b_59] 1 [b_60] 2 [b_61] 1 [b_62] 2 [b_63] 1
$
---------------------------------------------------------------------------
The even bits are used exactly twice while odd ones are only used once
(here odd and even both refer to the position). Or you could say that even
bits are duplicated in the internal state built after this step.
Anybody familiar with the DES knows that the IP(X) function of the DES
gives the internal state L || R where:
- L is an array composed of the odd bits of X
- R is an array composed of the even bits of X
In an academic WB DES implementation, building the 96 bits state is
performed using the duplication of even bits (R). This is because these
bits are necessary as both input of the E-box and output of the DES round
function (see my previous description of DES). So we have an obvious match
and it's a clear indication that there is no external encoding applied to
the input (and as a consequence probably none applied to the output as
well). More precisely there could still be a bit permutation on both L & R
bits but it sounds like a silly hypothesis so let's forget about that.
What would be the point?
---
Now let's continue with the differential analysis of the full wb_init().
This step is much more intuitive. Think about it: if you want to discover
the nibbles of stage0 (the output of wb_init) influenced by a specific
input bit then apply wb_init() to two inputs whose only difference is this
bit. Then calculate the XOR of both results and the non null nibbles are
the ones which are affected. This was greatly inspired by [R09].
---------------------------------------------------------------------------
$ ./entropy
[...]
[+] Differential cryptanalysis on wb_init()
-> b_00 :: 00 04 20 00 00 00 00 00 00 00 00 00
-> b_01 :: 00 00 00 40 00 00 00 00 00 00 00 00
-> b_02 :: 00 00 00 09 d0 00 00 00 00 00 00 00
-> b_03 :: 00 00 00 00 00 00 00 90 00 00 00 00
-> b_04 :: 00 00 00 00 00 0e 60 00 00 00 00 00
-> b_05 :: 00 00 00 00 00 00 00 00 00 50 00 00
-> b_06 :: 80 00 00 00 00 00 00 05 00 00 00 00
-> b_07 :: 00 00 00 00 00 00 00 00 00 00 00 b0
-> b_08 :: 00 07 00 00 00 00 00 00 01 00 00 00
-> b_09 :: 00 00 00 f0 00 00 00 00 00 00 00 00
-> b_10 :: 00 00 00 06 00 00 00 00 00 03 00 00
[...]
---------------------------------------------------------------------------
So for even bits there are 2 nibbles affected and only one for odd
bits. Not only does it confirm our previous hypothesis but it also reveals
the position (the index in the nibble array) of the bits in the WB internal
state (up to 1/2 probability for even bits). This is particularly
interesting when it comes to locate S-box for example ;-)
Analyzing the first wb_round()
------------------------------
To analyze this function, one clever trick is to make use of the odd
bits (L0) and perform a differential analysis.
Natively, the DES satisfies the following system of equations:
L1 = R0
R1 = L0 [+] f(R0,K0)
With
L0 || R0 being the result of IP(plaintext)
K0 being the first subkey
Let's now consider two plaintexts (A and B). The first one is composed
of bits all set to 0 (L0_A || R0_A) whereas the second one ((L0_B || R0_B)
has a weight of 1 and more specifically, its sole bit set to 1 is in L0.
Remark: While there is only one A, there are obviously 32 possible B.
We can thus write thanks to the previous equations:
L1_A = R0_A = 0
R1_A = L0_A [+] f(R0_A,K0) = f(0,K0)
And
L1_B = R0_B = 0
R1_B = L0_B [+] f(R0_B,K0) = L0_B [+] f(0,K0)
(Again please excuse the lazy notation)
This finally gives us:
DELTA(L1||R1)(A,B) = ( L1_A [+] L1_B || R1_A [+] R1_B )
= ( 0 [+] 0 || f(0,K0) [+] L0_B [+] f(0,K0) )
= ( 0 || L0_B )
We know that L0_B's weight is 1 so in a native DES the modification of
one bit in L0 induces the modification of a unique bit in the output of the
DES round function. In an obfuscated context, this means that only one
output nibble is modified and calculating DELTA (the result of the so
called differential analysis if you prefer) is merely a trick to identify
it easily.
Now that you've grasped the main idea, let's work on the real WB. Again
consider plaintexts A and B which give (L0_A || R0_A) and (L0_B || R0_B)
after IP().
Because wb_round() includes the E-box and produces a 96 bits output
state, we now have to consider an additional transformation:
X (64b) ---> [ wb_init + first wb_round ] ----> Y (96b)
Here Y is the output of wb_round. Following the design in academic
publications we can write:
Y = RP ( L1 || X1 || r1 ) (RP = Random bit Permutation used to hide
the position of bits in the
obfuscated output.)
With:
- L1 being R0 (from DES round equation)
- X1 being the result of the E-box applied to R1
- r1 being the complementary bits such as the set of X1 and r1 is
exactly twice R1
Now let's apply again the differential analysis. It's important to
remark that RP() and E() are both linear operations as this simplifies
things. Indeed it's well known that:
LinearFunc(x [+] y) = LinearFunc(x) [+] LinearFunc(y)
Putting everything together this gives us:
DELTA(Y)(a,b) = RP(Y_A) [+] RP(Y_B)
= RP(Y_A [+] Y_B)
= RP(L1_A [+] L1_B || X1_A [+] X1_B
|| r1_A [+] r1_B)
= RP(0 [+] 0 || E(f(0,K0)) [+] E(L0_B [+] f(0,K0))
|| r1_a [+] r1_b)
= RP(0 || E(f(0,K0) [+] L0_B [+] f(0,K0)z)
|| r1_A [+] r1_B)
= RP(0 || E(L0_B) || r1_A [+] r1_B)
If the bit set in L0 is a middle bit then:
- Weight(E(L0_B)) = 1 and Weight(r1_A [+] r1_B)) = 1
If the bit set in L0 isn't a middle bit then:
- Weight(E(L0_B)) = 2 and Weight(r1_A [+] r1_B)) = 0
In both cases, Weight(RP(0 || E(L0_B) || r1_A [+] r1_B)) = 2, RP having
no effect on the weight since it only permutes bits. This means that 1 bit
modification should have a visible impact on 'at most' 2 nibbles. 'at most'
and not 'exactly' because with the effect of RP() the two bits could be
located in the same nibble.
Let's see if we are right:
---------------------------------------------------------------------------
b_01 :: 00 05 d0 00 00 00 00 00 00 00 00 00 <-- 2 modified nibbles
b_03 :: 00 00 00 03 60 00 00 00 00 00 00 00 <-- 2 modified nibbles
b_05 :: 00 00 00 00 00 04 e0 00 00 00 00 00 <-- 2 modified nibbles
b_07 :: 90 00 00 00 00 00 00 08 00 00 00 00 ...
b_09 :: 00 0b 00 00 00 00 00 00 05 00 00 00
b_11 :: 00 00 00 0f 00 00 00 00 00 08 00 00
b_13 :: 00 00 00 00 00 0d 00 00 00 00 0f 00
b_15 :: 00 00 00 00 00 00 00 0f 00 00 00 06
b_17 :: 00 04 00 00 00 00 00 00 0c 00 00 00
b_19 :: 00 00 00 09 00 00 00 00 00 0f 00 00
b_21 :: 00 00 00 00 00 08 00 00 00 00 06 00
b_23 :: 00 00 00 00 00 00 00 0d 00 00 00 08
b_25 :: 08 d0 00 00 00 00 00 00 00 00 00 00
b_27 :: 00 00 04 20 00 00 00 00 00 00 00 00
b_29 :: 00 00 00 00 05 80 00 00 00 00 00 00
b_31 :: 00 00 00 00 00 00 04 20 00 00 00 00
b_33 :: 02 70 00 00 00 00 00 00 00 00 00 00
b_35 :: 00 00 0c f0 00 00 00 00 00 00 00 00
b_37 :: 00 00 00 00 0d b0 00 00 00 00 00 00
b_39 :: 00 00 00 00 00 00 0f a0 00 00 00 00
b_41 :: 0c 00 00 00 00 00 00 00 0f 00 00 00
b_43 :: 00 00 0d 00 00 00 00 00 00 02 00 00
b_45 :: 00 00 00 00 09 00 00 00 00 00 05 00
b_47 :: 00 00 00 00 00 00 03 00 00 00 00 03
b_49 :: 0f 00 00 00 00 00 00 00 0d 00 00 00
b_51 :: 00 00 06 00 00 00 00 00 00 03 00 00
b_53 :: 00 00 00 00 0b 00 00 00 00 00 0c 00
b_55 :: 00 00 00 00 00 00 02 00 00 00 00 01
b_57 :: b0 00 00 00 00 00 00 0c 00 00 00 00
b_59 :: 00 03 60 00 00 00 00 00 00 00 00 00
b_61 :: 00 00 00 0e 40 00 00 00 00 00 00 00
b_63 :: 00 00 00 00 00 0b f0 00 00 00 00 00
---------------------------------------------------------------------------
And that's exactly what we were expecting :) Well to be honest, I first
observed the result of the differential analysis, then remarked a 'strange'
behavior related to the odd bits and finally figured out why using maths ;)
One cool thing with this situation is that we can easily leak the
position of the specific S-Boxes inside the T-Boxes. First let's compare
the differential analysis of even bits 28,36,52,60 and of odd bit 1:
---------------------------------------------------------------------------
b_01 :: 00 05 d0 00 00 00 00 00 00 00 00 00
b_28 :: 0d 75 dd 00 00 00 04 20 0f d2 00 00
b_36 :: 0c 05 d0 00 09 00 04 20 cf 00 05 00
b_52 :: 00 05 d0 09 00 00 00 00 90 0f 00 00
b_60 :: 0c 05 d6 09 00 00 02 00 3f 0d 00 01
---------------------------------------------------------------------------
Obviously setting these even bits one by one induces the same
modification (amongst others) as setting the odd bit 1 (nibbles 01L (0x5)
and 02H (0xd)) so there must be some kind of mathematical link between them
because the other bits do not have such property.
Playing with Sbox
------------------
The reason behind this behavior is very simple to explain. But first,
let's take back the example of plaintext 'A' (null vector):
We know that:
R1_A = L0_a [+] P(S1[0 [+] k0] || S2[0 [+] k1] || ... || S8[0 [+] k7])
R1_A = 0 [+] P(S1[k0] || S2[k0] || ... || S8[k7])
R1_A = P( S1[k0] || S2[k1] || ... || S8[k7] )
Where:
The ki being 6 bits vectors (0 <= i < 8)
K0 = k0 || k1 || k2 ... || k7
Thus in the case of plaintext 0 (A), R1_A is the permutation of the
Sbox output whose inputs are the bits of the first subkey.
Now let us focus on 1 of the 4 bits generated by an Sbox S (which could
be any of the 8). We do not know its value (b) but when the P-box is
applied it will be located in a particular nibble as illustrated below:
R1_A = f(R0,K0) = ???? ?b?? ???? ???? ???? ???? ???? ????
^
|__ The bit
<------------------------------------->
(4bits x 8) = 32 bits state
Because a WB DES implementation is working with a duplicated Rx this
will give us the following internal state:
... ??b? ???? ???? ???b ...
^ ^
| |
-------------------- b is duplicated
<------------------------->
96 bits state
Now following what was explained previously with odd bits, out of the
32 possible B, one of them will affect b when L0_B is XORed with f(0,K0)
So considering a 96 bits internal state inside the WB, this gives us:
... ??a? ???? ???? ???a ...
With:
a = b [+] 1
As a result, the differential between A and B would be:
... ??b? ???? ???? ???b ... (from A)
[+]
... ??a? ???? ???? ???a ... (from B)
=
... ??1? ???? ???? ???1 ... ( because a [+] b
= a [+] a [+] 1
= 1 )
From now on, we will call this differential our 'witness' and by
extension, the two nibbles where b=1 the 2 witness nibbles.
Playing with the witness
------------------------
Now imagine that we're using another plaintext (X) with weight 1 and
whose weight is in one of the 6 possible bits influencing Sbox S. There are
two possible situations:
- S still produces b
- S now produces b+1
If we perform a differential analysis between X and A (null vector)
this gives us:
case 1:
=======
... ??b? ???? ???? ???b ... (from A)
[+]
... ??b? ???? ???? ???b ... (from X)
=
... ??0? ???? ???? ???0 ... <-- useless output
case 2:
=======
... ??b? ???? ???? ???b ... (from A)
[+]
... ??a? ???? ???? ???a ... (from X)
=
... ??1? ???? ???? ???1 ... <-- witness vector :)))
So case 2 is perfect because it gives us a distinguisher. We can test
all 32 possible X (each of them having a different even bit set) and
observe the ones which produce the witness vector associated with b.
This is exactly what we did implicitly when we discovered the link
between bits 28, 36, 52 and 60. Or if you're lost let's say that we've just
discovered something huge: the bits 28, 36, 52 and 60 are the input of the
same Sbox and bit 1 is one of the output of this Sbox. At this point the
protection took a heavy hit.
Remark: The first subkey is modifying the input sent to the Sbox. As a
consequence the relation previously found is "key dependent". This will be
of importance later, keep reading!
Going further
-------------
Let's think. At this point and thanks to our analysis of wb_init()
we're almost sure that there is no external encoding applied to the input.
So there should be a match between our practical results and the
theoretical relations in the original DES algorithm. To verify my theory, I
wrote a little script to compute the positions of the bits involved with
each Sbox:
---------------------------------------------------------------------------
$ ./bitmapping.py
[6, 56, 48, 40, 32, 24] <-- Sbox 1
[32, 24, 16, 8, 0, 58] <-- Sbox 2
[0, 58, 50, 42, 34, 26]
[34, 26, 18, 10, 2, 60]
[2, 60, 52, 44, 36, 28] <-- Sbox 5
[36, 28, 20, 12, 4, 62]
[4, 62, 54, 46, 38, 30]
[38, 30, 22, 14, 6, 56] <-- Sbox 8
---------------------------------------------------------------------------
Oh interesting so Sbox 5 seems to match with our practical result.
Going deeper, we need to check if bit 01 is involved with this Sbox. Again
I wrote another script to compute the position of odd bits involved with
the Sbox in the original DES and this gives us:
---------------------------------------------------------------------------
$ ./sbox.py | grep 'SBOX 5'
bit 41 XORED with bit 00 of SBOX 5 (19)
bit 01 XORED with bit 03 of SBOX 5 (16)
bit 19 XORED with bit 02 of SBOX 5 (17)
bit 63 XORED with bit 01 of SBOX 5 (18)
---------------------------------------------------------------------------
So bit 01 is indeed involved. However let's try to be careful. In
cryptanalysis it's easy to be fooled, so let's make extra checks. For
example can we link a subset of even bits {2, 28, 36, 44, 52, 60} with bit
19 of the same Sbox?
---------------------------------------------------------------------------
19 :: 00 00 00 09 00 00 00 00 00 0f 00 00
2 :: 0c 00 06 00 00 0b f2 60 0f 03 00 01
28 :: 0d 75 dd 00 00 00 04 20 0f d2 00 00
36 :: 0c 05 d0 00 09 00 04 20 cf 00 05 00
44 :: 00 00 00 09 00 0b f0 00 20 0f 00 00
52 :: 00 05 d0 09 00 00 00 00 90 0f 00 00
60 :: 0c 05 d6 09 00 00 02 00 3f 0d 00 01
---------------------------------------------------------------------------
Bit 19 is linked to bit 44 and 52 => YES. At this point, we should
check automatically that the bit relations are satisfied for all the Sbox
but it's tedious. That's the problem :-P Because I was lazy, I manually
checked all the relations. Fortunately with the help of scripts, this only
took me a couple of minutes and it was a 100% match. Again, this proves
nothing but as I said earlier, we're working with guesses.
Towards a perfect understanding of differential analysis
--------------------------------------------------------
Didn't you notice something particular with bit 02, 28 and 60? Well the
'impacted' nibbles were neither 0 nor a witness nibble. For example
consider bit 60:
---------------------------------------------------------------------------
19 :: 00 00 00 09 00 00 00 00 00 0f 00 00
60 :: 0c 05 d6 09 00 00 02 00 3f 0d 00 01
---------------------------------------------------------------------------
The first impacted nibble '0x9' is a good one (witness nibble) but the
second one is neither '0x0' nor '0xf' (witness). How is that possible?
Well the answer lies in both:
- the (non)-middle bits
- the P-box
Indeed if you consider the bits sent to Sbox 5, you have to know that:
- bits 02 and 60 are sent to both Sbox 4 & 5
- bits 52 and 44 are sent to Sbox 5
- bits 36 and 28 are sent to both Sbox 5 & 6
So when 1 non-middle bit is set, this will impact the output of 2 Sbox
and we're unlucky, the P-box will have the unfortunate effect of setting
them in the same nibble, hence the difference observed.
----[ 6.3 - Recovering the first subkey
If the relations observed are 'key dependent', considering the fact
that the S-Boxes are known (which means unmodified otherwise this would be
cheating :p) then isn't this an indirect leak on the key itself that could
be transformed in a key recovery? Oh yes it is :-)
First cryptanalysis
-------------------
The main idea is really simple: we know that for a given subkey,
several unitary vectors (plaintexts of weight 1) will produce the same
output bit.
Let's take again the previous case. We have:
.------.------.------.-----.-----.------.
| b_02 | b_60 | b_52 |b_44 |b_36 | b_28 |
'------'------'------'-----'-----'------'
.....
. + .
.....
.------.------.------.-----.-----.------.
| k24 | k25 | k26 | k27 | k28 | k29 |
'------'------'------'-----'-----'------'
|
v
*********************
* Sbox 5 *
*********************
|
v
.------.------.------.-----.
| y0 | y1 | y2 | y3 |
'------'------'------'-----'
Let us consider bit 01. We know that it will be XORed to y2 so from the
differential analysis we can derive the set of relations:
[ k24 [+] 0, k25 [+] 1, k26 [+] 0, k27 [+] 0, k28 [+] 0, k29 [+] 0 ] => b
[ k24 [+] 0, k25 [+] 0, k26 [+] 1, k27 [+] 0, k28 [+] 0, k29 [+] 0 ] => b
[ k24 [+] 0, k25 [+] 0, k26 [+] 0, k27 [+] 0, k28 [+] 1, k29 [+] 0 ] => b
[ k24 [+] 0, k25 [+] 0, k26 [+] 0, k27 [+] 0, k28 [+] 0, k29 [+] 1 ] => b
So amongst all possible sets {k24,k25,k26,k27,k28,k29}, only a few of
them (including the one from the real subkey) will satisfy the relations.
Testing all possible sets (there are 2^6 = 64 of them) will give us 2 lists
because we do not know if b=1 or b=0 so we have to consider both cases.
Applying this technique to both y0, y1, y2 and y3 will allow to filter
efficiently the number of possible candidates as we will only consider
those present in all lists. The success of this cryptanalysis is highly
dependent on the number of relations that we will be able to create for a
particular S-Box. Practically speaking, this is sufficient to recover the
first subkey as the complexity should be far below 2^48. Should be? Yes I
didn't test it... I found even better.
Immediate subkey recovery
-------------------------
As I said above, our success is dependent of the number of equations so
improving the cryptanalysis can be done by finding ways to increase this
number. There are two obvious ways to do that:
- There may exist combinations of input bits other than unitary
vectors (weight > 1) which can produce the witness nibbles in a
differential analysis.
- If the impacted nibbles are both 0x0 then this gives us a new
relation where expected output bit is b [+] 1
Practically speaking this gives us the following result for Sbox5 and
bit 01:
---------------------------------------------------------------------------
$ ./exploit
[...]
{ 1 0 0 0 0 0 } = { 1 } <-- dumping relations for S5 & bit 01
{ 0 1 0 0 0 0 } = { 0 }
{ 1 1 0 0 0 0 } = { 0 }
{ 0 0 1 0 0 0 } = { 0 }
{ 1 0 1 0 0 0 } = { 1 }
{ 0 1 1 0 0 0 } = { 1 }
{ 1 1 1 0 0 0 } = { 1 }
{ 0 0 0 1 0 0 } = { 1 }
{ 1 0 0 1 0 0 } = { 0 }
{ 0 1 0 1 0 0 } = { 0 }
{ 1 1 0 1 0 0 } = { 1 }
{ 0 0 1 1 0 0 } = { 1 }
{ 1 0 1 1 0 0 } = { 0 }
{ 0 1 1 1 0 0 } = { 1 }
{ 1 1 1 1 0 0 } = { 0 }
{ 0 0 0 0 1 0 } = { 0 }
{ 1 0 0 0 1 0 } = { 1 }
{ 0 1 0 0 1 0 } = { 0 }
{ 1 1 0 0 1 0 } = { 0 }
{ 0 0 1 0 1 0 } = { 1 }
{ 1 0 1 0 1 0 } = { 0 }
{ 0 1 1 0 1 0 } = { 0 }
{ 1 1 1 0 1 0 } = { 1 }
{ 0 0 0 1 1 0 } = { 0 }
{ 1 0 0 1 1 0 } = { 0 }
{ 0 1 0 1 1 0 } = { 1 }
{ 1 1 0 1 1 0 } = { 0 }
{ 0 1 0 1 0 1 } = { 1 }
[...]
[ key candidate is 31]
---------------------------------------------------------------------------
The cryptanalysts have the habit to always evaluate the complexity of
their attacks but in this case let's say that it's useless. Only one subkey
appeared to be valid out of the 2^48 possible ones.
----[ 6.4 - Recovering the original key
Now that we've retrieved the first subkey, our goal is almost reached.
So how do we retrieve the secret key? Well DES subkeys can be seen as
truncated permutations of the original key. This means that we now have 48
out of the 56 bits of the original key.
I could explain the key scheduling mechanism of the DES, but it's
useless as the only important thing is to be able to reverse the
permutation. This is done easily thanks to the following python
manipulation applied to the sKMap1 array, itself being shamelessly ripped
from [13]:
---------------------------------------------------------------------------
>>> InvsKMap1 = [ -1 for i in xrange(64) ]
>>> for x in xrange(len(InvsKMap1)):
... if 7-x%8 == 0:
... InvsKMap1[x] = -2
...
>>> for x in xrange(64):
... if x in sKMap1:
... InvsKMap1[x] = sKMap1.index(x)
...
>>> InvsKMap1
[19, 8, 12, 29, 32, -1, -1, -2, 9, 0, -1, -1, 44, 43, 40, -2, 5, 22, 10,
41, 37, 24, 34, -2, 15, 14, 21, 25, 35, 31, 47, -2, 6, 2, 13, 20, 28, 38,
26, -2, 23, 11, -1, 16, 42, -1, 30, -2, 4, -1, 1, -1, 33, 27, 46, -2, 7,
17, 18, 3, 36, 45, 39, -2]
>>>
---------------------------------------------------------------------------
Here is the resulting array:
char InvsKMap1[64] = {
19, 8, 12, 29, 32, -1, -1, -2,
9, 0, -1, -1, 44, 43, 40, -2,
5, 22, 10, 41, 37, 24, 34, -2,
15, 14, 21, 25, 35, 31, 47, -2,
6, 2, 13, 20, 28, 38, 26, -2,
23, 11, -1, 16, 42, -1, 30, -2,
4, -1, 1, -1, 33, 27, 46, -2,
7, 17, 18, 3, 36, 45, 39, -2
};
My exploit uses this array to build an original key out of both the
subkey bits and an 8 bits vector. '-1' is set for a bit position where the
value has to be guessed. There are 8 such positions, and for each of them,
a bit is taken from the 8 bits vector. '-2' means that the bit can be
anything. Indeed the most significant bits (the so-called parity bits) of
the 8 bytes key array are never taken into account (hence the well known
8 x 7 = 56 bits keylength).
Now the only remaining thing to do is to guess these 8 missing bits.
Obviously for each guess you will generate an original key 'K' and test it
against a known couple of input/output generated by the white-box. The
whole operation was implemented below:
---------------------------------------------------------------------------
void RebuildKeyFromSk1(uchar *dst, uchar *src, uchar lastbits)
{
int i,j;
char *plastbits = (char *)&lastbits;
memset(dst, 0, DES_KEY_LENGTH);
for(i=0,j=0; i<64; i++)
{
// Parity bit
if(InvsKMap1[i] == -2)
continue;
// Bit is guessed
else if(InvsKMap1[i] == -1)
{
if(GETBIT(plastbits,j))
SETBIT(dst,i);
j++;
}
// Bit is already known
else
{
if(GETBIT(src, InvsKMap1[i]))
SETBIT(dst,i);
}
}
return;
}
[...]
const_DES_cblock in = "\x12\x32\xe7\xd3\x0f\xf1\x29\xb3";
const_DES_cblock expected = "\xa1\x6b\xd2\xeb\xbf\xe1\xd1\xc2";
DES_cblock key;
DES_cblock out;
DES_key_schedule ks;
for(missing_bits=0; missing_bits<256; missing_bits++)
{
RebuildKeyFromSk1(key, sk, missing_bits);
memset(out, 0, sizeof out);
DES_set_key(&key, &ks);
DES_ecb_encrypt(&in, &out, &ks, DES_ENCRYPT);
if(!memcmp(out,expected,DES_BLOCK_LENGTH))
{
printf("[+] Key was found!\n");
[...]
}
}
---------------------------------------------------------------------------
The whole cryptanalysis of the white-box is very effective and allows
us to retrieve a key in a few ms. More precisely it retrieves _1_ of the
256 possible 8 bytes key ;)
---------------------------------------------------------------------------
$ tar xfz p68-exploit.tgz; cd p68-exploit
$ wget http://homes.esat.kuleuven.be/~bwyseur/research/wbDES
$ md5sum wbDES
b9c4c69b08e12f577c91ec186edc5355 wbDES # you can never be sure ;-)
$ for f in scripts/*.gdb; do gdb -x $f; done > /dev/null # is quite long
$ make
gcc -c wb_init.c -O3 -Wall
gcc -c wb_round.c -O3 -Wall
gcc -c wb_final.c -O3 -Wall
gcc exploit.c *.o -O3 -Wall -o exploit -lm -lcrypto
gcc wb_main.c *.o -O3 -Wall -o wbdes.try
gcc entropy.c -o entropy -lm
$ ./exploit
[+] Number of possible candidates = 256
-> Required computation is 2^(8) * DES()
[+] Key was found!
-> Missing bits: 0x3d
-> Key: '02424626'
$
---------------------------------------------------------------------------
And that's it! So the key was bf-able after all ;>
--[ 7 - Conclusion
Nowadays there are a lot of white-box protections in the wild (DRM but
not only) using either academic designs or their improvements. Each of them
is an interesting challenge which is why you may want to face it one day.
This paper is not ground breaking nor even relevant for the average
cryptographer, the cryptanalysis of the naked DES being covered in many
papers including [R16]. I wrote it however with the hope that it would give
you an overview of what practical white-box cracking could be. I hope you
enjoyed it :)
Feel free to contact me for any question related to this paper using
the mail alias provided in the title of the paper.
--[ 8 - Gr33tz
Many (randomly ordered) thanks to:
- the #f4lst4ff crypt0/b33r team for introducing me to the concept of
white-box a few years ago.
- Jb & Brecht for their implementations which gave me a lot of fun :)
- X, Y, Z who will remain anonymous but nonetheless helped me to
improve _significantly_ the paper. If you managed to understand a few
things out of this "blabla" then you must thank them (and especially
X). I owe you big time man :)
- asciio authors because without this tool I would never have found the
courage to write the paper
- The Phrack Staff for publishing it
--[ 9 - References
[R01] http://en.wikipedia.org/wiki/Feistel_cipher
[R02] http://2009.hack.lu/index.php/ReverseChallenge
[R03] http://baboon.rce.free.fr/index.php?post/2009/11/20/
HackLu-Reverse-Challenge
[R04] http://www.whiteboxcrypto.com
[R05] "Cryptanalysis of a White Box AES Implementation", Billet et al.
http://bo.blackowl.org/papers/waes.pdf
[R06] "Digital content protection: How to crack DRM and make them more
resistant", Jean-Baptiste Bedrune
http://esec-lab.sogeti.com/dotclear/public/publications/
10-hitbkl-drm.pdf
[R07] "White-Box Cryptography and an AES Implementation", Eisen et al.
http://www.scs.carleton.ca/%7Epaulv/papers/whiteaes.lncs.ps
[R08] "White-Box Cryptography and SPN ciphers", Schelkunov
http://eprint.iacr.org/2010/419.pdf
[R09] "A White-box DES Implementation for DRM Applications", Chow et al.
http://www.scs.carleton.ca/%7Epaulv/papers/whitedes1.ps
[R10] "White-Box Cryptography", James Muir, Irdeto
http://www.mitacs.ca/events/images/stories/focusperiods/
security-presentations/jmuir-mitacs-white-box-cryptography.pdf
[R11] http://search.cpan.org/dist/App-Asciio/lib/App/Asciio.pm#NAME
[R12] http://dhost.info/pasjagor/des/start.php
[R13] "Cryptography: Theory and Practice", D. Stinson, 1st edition
[R14] "Clarifying Obfuscation: Improving the Security of White-Box
Encoding", Link et al.
http://eprint.iacr.org/2004/025.pdf
[R15] "White-Box Cryptography" (PhD thesis), B. Wyseur
https://www.cosic.esat.kuleuven.be/publications/thesis-152.pdf
[R16] "Attacking an obfuscated cipher by injecting faults", Jacob et al.
http://www.cs.princeton.edu/~mjacob/papers/drm1.pdf
[R17] "Cryptanalysis of White-Box DES Implementations with Arbitrary
External Encodings", B. Wyseur
http://eprint.iacr.org/2007/104.pdf
--[ 10 - Appendix
begin 644 p68-exploit.tgz
M'XL(``$N74\``^Q<>W,:Q[+WO_`I)DY9`1G)^P9)H)0?<J*R8[DLY22G$*$6
M6*05CR7L8J/CH_O9;W?/8V<?2,B5W%OW015>=J:GNZ>GIZ?G-R,OO-9>L%Y,
MHS!Y\>1O^ACP:;HN/LVF:^I/^7EBFDVK:5E>T_&>&/#BF$^8^W<II']6<>(O
M&7L2W\:3^^@>JO\?^EEHX_^+/PG&X33XJV7@`'N.LW'\7<^C\?>:INDUP4],
MVS&:3YCQ5RM2]OD_/O[^='K(HL%-S(07L"^#41`GRUL6S)-EM+BM5K'ZL%JY
M&@[9WA#J^^$\3/;AY<QF>[\!![UN&:WFHTV5XW#N3S.552'V4,K?'_(&ZI7M
M[D=I`[87*4WWIC/X#I>WBR2J5J7>ARAHYH=SR4F]%CE1FWUH!'KPWA[*;BL]
MY"M)YB\HN3J<!OX<S+*<$5]9I5A*-:O+P)]FB*^K_]VCGG[T^:],_A?+N'_^
M.YZ1SG_7L1R<_QXL"?\___\+/M^'\^%T-0I8.TY&8;1_?5S-%$W#0:$LG"?Y
MLF4XO\J6#9/;19`MFOG)=;8D6@3S.)Z^P"D#%=47N^P7?[B,8K;[HEI%#J-@
MS%;S.+R:!R,VO(:A6N&_1]7J]U`5S@/VYN2\?_KAXZ\7_7^<O+XX^]0W*D\O
MUX9YN;;LR[7C7JZ]YN6Z=7"Y?OGJ<OWZS>7ZY.W33.NS7R_RS5O0+&A=KDU@
MX3K`;@Q?_W(]P-]NMOF[DW]FVF(;&^A<$-L$L0<@]M7KR_6;MY?KMV;:]K=7
M)7J;%C2&;P"-1S87/,:^`*.!G6E<HK8/E-X`6B('>`Z@=0!E(_@.K:>IT5Z_
M__3J]*+&SB\^-=CIF]]9G=58#=[JW1J\UE^T>FRGP_ZC9H`E6;O-:DVVQVI4
M]ZQ5K]=9/<?+>@PSIAAI?,Y/+A[4Z=\=MHU*G-4#*F5XE2GT4XE"P"7+ILZ.
MC_.J`.$.0]9Y7M96S%(^G`NHE'&X5^_/7K_KOS_Y\-/%SY56P1=%1:[FPZ?^
M^:NSW_5B\*'SBY<7)ZJ%:=$4_+B,D@@G'Y^&-.'4ZLT7^!HOW&TP"`;UH^KG
M*!S)S""MVT"4XY"CXAK`;P@I;+R:#Y,PFI,BU'ZTFBWZ@]5X'"QKO.%B"=U9
M*SY01ZQ8'/XKJ%>_5BOX$G8,8%T)QS5.7J]6*@N4,JX]?1:SIPU1C$3C:%D[
M8F$;&<#S^?,,\;ZU!G(&8KIA#^A5S>7\*;XN@V2UG!]5[Z@G8'IVOC>(UAEC
MGK^*UEUM5'I=S^FQ#BC+X/.5,6AY;F`#?#>=!F,&?.%INO"UX7<3OB8O8U:#
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MM@NCP_LH+`FAKF@E6UJ)*&RMA[(GTM[Z+')3K<78IE;2),I9DO%^.9.D!=08
M-M1HD<R<'"E?MI6\<GJG\^*@D>HC+2/[)/0IZ$GZE%C)45;2[&WJO97CWQ1<
MA*^IN.2E?D#2W$8FYM"L<!NIKTI_/=!F%??5-"[)6=W2K"HM8FBS6$8#06O:
MFI5$W#*E163\.=#\5,Y6W:?*XI*K9IRE:>!J8RT\EK@+C51\.6A(RZB80AK(
M62/'6/J,U*BI17-;Q1`5VU0\M#7Y^OBG<KG>DH^<^9*_C!AR]LEH8FIZ&HUL
M#"NSDJ=\2<Q_R5VM8V*\2!/92ZEE&IG3-4Z?_W(M:VHQ2LX8.=XMI74Z:[V&
M\BGILYE8)R.!U,U0>J:S5OB1OGY1>U>SGNX!MO2`$BLUE2_)\=<BJEH_I)UE
M[Z0UTTPA70&D5+EF>MEX(&.,LF;JH\H/F!;#E%])>=+7+,TZ:9:0KI/".FJ.
M'&B^(W63,5<;,=/D5JK>0;Y&&54,*58_^!S,^X,PB7.Y%:16#/+`K]3>A:\#
M/!R088-<RP%KLQ<O()5`$EX$(J1S`)G;DB06<:$B^,+3`7(;R"U/DMC$A8J`
M2TMT#,@\0Y(XQ(6*@`L\'2"W@=Q2@ESB0D7PE4,&9)XE23SB0D7`!9X.D-M`
M;BM!3>)"1<!%IFQD!';'B*15K62-&(U&]]I0YBP>C+5[`)(JRG8N%#LPA#9\
MK::L(IN9X!66&DU91;9RH<H%9C8^5179R(%B#_V8UB]91;9Q0`T'JFQL:<LJ
M;A,4C\4T]V45V<(ZX$[KH#H@JZ*9`-/U?P3#)%JR93#UDV#$X@0V&I2WQ\ER
M-4S89UX/AN")_&+JA_-N?E/6.T*^5)<$ZT02Q\F5T<UMNCBIW#G5-5)S(RG?
M/]7)^;-Z\0?L=,J*3=QD'%6E`/\J,/LS/YX$(ZPH2A,;KL$JG([Z*U#/7][V
M.:NXAE7IYJHQPWW3+)C%05+;$5HT<)[@'BH:2\5P?U3%7IP"N]"?0J6L$ONN
M&>S3V*R=4P:*^"Y,&DHTVB<;-^0;&K@Q`R&LHLR4J=->S(9!F[VL,LEUP#R'
MB=X*W>*T;\*2Q:Z9V#7L`&XT8>_H.6KG*'`-.0AA3VC-0FZ.TE9@V6WLD3.(
MQETKT,R2MTM:S[(%W#R5.VZA5^@#9!SN,,Q?+OW;U"P9;]*-DZG8LKO%WLK.
M9ITV['5G&)5RBF/A'TP?:2BAKNA;\^P$H85##/;&66)M,4UT1C4"(3"DAJ,U
MS17.-`1E%*2)-#>9Z:.ST$VIERN88J,A;[#BINW!/]QX7RMH08&8A&`V0D]V
MZSNH3@6!$0E1+<+&39WL+5TW8Y[4?W-KKNQJKWM#N`A:_.XA/;?U[PTJY(NY
M*W-?S[A[*6%)H>;W9;YM;71N:^.8,/'Y!A^WE).7J;J=IZ>HV0/.J;MDB)U1
MP%:R=\S>``=$Y#R;=)&AD4&/&*8([-GH<OZTH5BJ<&@6!SYE_&S?&K'#0_:T
M0<&PHJ-['WY]_[Z1GX)AKY''+?EXW0_#?0@'@VD0;U[8YT3`=#O07/GL3X_$
M^$$<'"^C&>0A2<3>Y@D'MTG07T3Q44IH("&DJCG*:#P&OSH2+`VRX/OH"WO!
M'/K]<WAU36L[Z/U;\,,R8-,HFA`:"K46BQ?!,!R'0Z%S#`TA9V,P!ZE#$MV%
MU.WLS9L^S-]S2&IS_?3!PD,P1%_PZ%J[&CWE>E^-NSL9W'`*]O-M2G(`^B5$
M=)B1S@CS*&QW//LH?-ZQM@WW7T5<JM5*XO[QL5/?,=;C.OL.!-',(?I*H6?\
MV=N'D<2H=R^SH_M9R#$&/K,'2/D@`Z'#"9\_YS7T=B=[5J++-W5J$Y^_H3_&
MAO[<;8X[FQTG&W,:-WK4Z3[OJ:`C.4B7%W-<B[::]Y9'FE<P/9Y9HT,,4LS:
M#9]3QJ:OE5:Z5F*GTJ80IN+\Q`&-:<I#V&/=9_OFNH>,RVV(TFZD"7]D3W\^
M_>GGIPR"WONSWQYH(T?H?BKP@G3!O3\0OJ;;$BH.RO.5^)`MP\4"2S!R_?CC
MC^SP.#UT^1@L9ZLDD&<VHSA1QRWQ<JA^S_P%/WJ!'YG3%XA-0UB&$Q$A0I%X
MOYX&T`SO8F`"@2;%%!.O<81S']7:5ZLPB<2E77!&#I(KNB\O9KNLE?4+)5ES
M"I7KU$AY:(LG.33=1,)#TL)Z82U%Y7&S%\;]:)*Z,73E0Y0$APSZ$^/!T]4<
M`[3/>PX!GTV"6W*7F!%U"/X4)6P(A@]'P3+`B1+@:N%Z`Z3=A]YA9=)':</!
M-!I.\#X)=[U.V:DUZ*K1HKQ.X73ZJ(0K&+U3<H:>91>M$E$`C/OQ\#H8K6`R
M3&(QH&`B-#?:X>P=XX,+U+!'50,(+/3<"5[K@B74(MO:#OS38#N36%8$PT$_
MF-/]GMH.9GL[Q`0H&J3QR8?7G_[Y\:(N#O:^`TG#V8(D25LU\EMS&N8E13*3
M%#Y[AYL2/L;8CS1VP;(DI6<7\%W4)5>$_<E$-"I.@MFB9,?`;89Y:7\^6-+"
M22;[[16MM+@7C<G_Y7YIRUTQZ(6=G\F-[?L@^2'&#&J9T,SB5Z5X\B.69J4%
M2E`O;=/5WM3$00,O;FO8JP:CL2BF8C+;35-O(D=JQ5!,+-+QY1AFRC4L7XR.
MEYD/B247S0LT7&#6SH^F3!7(SI0GYXZII3DJ5:D\EI!&!<\HQ,K3,5EM"?GN
M,@Y8.&;CJ?\%-[A?`C8/8)#!F*,@`4]C(OG"@06;Y.*#V/?-NZAAX=K'D220
M3BO)BC.<TT$72B"GDIE8%5[5WVHN:OZ.#LY+_^J9)?;5[W[Q%V;7:0E4\8"#
MJ;:-D"*'9Q%_12P6X5JS0:`O@K8FK(02<G4Y4HLH(&&(+>H4HGX$O38)XP5R
M$]%(#HTB0HFHJLW14H0#'0[!(CIJM8`<,5D$7!&.].0)UP'':.DDP!+8IT-(
MJP:?GLX_BX[1:3YVS#Q(CPU0&/9HSQ1?^%VMR",]6>J('B$\+2A<@=T:U']2
M#/N--N`4ICB(HIZZW`2V0#D%#T^![]0Y1)71'F@H3F&)XP+4`6WO"$WM5`^'
MES!!A:.%=G44#SH,:?*18L+2B-,B'KQG<4O!O'H9QZL9IG7)M9_@7+J&24]3
M;1PN<?U<#6@I@`D'E;",LL4R^@P+)087))OB*KL,\%8GLFGQ]`%JYC`/(>)A
M1G$51;"SK*<YS*>`()IWP>U;2''.)^8#V0P*0;[:#N<&%U".H,A:":2PW?J.
M+-/`')FV9._F:+EKXZ8,&8'9\]%?ALDM]HQG+*ESA>!;';`G)2RPGD.RM`I0
M)B%U$(@@N[A:!7$<C*`LF%+H*K8W]3V6R(=4M_(($/:#[]`KD".+7%.3YT^7
M@3^Z99-Y]&4NI!;9DW5U138)*6PFT/Z%`<R'V9+<)IO&R-BU*;E)DZ="09\L
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M#,(K*L7<HU(90%":/+`QSN;\Y*`<1MEPU$6SK0RC$"<CHE41H=@(@4H$E(6L
M'`/-HSN-QP*@`ON+;V/(?)GH">3A_>#/-'0%?ZYHSQ%W/;O7]9Z;$,6T$T;1
M^/PVOL#4QE#8((>6_\PBRAP&@"%"\$*_A$GKW09H&4=;:B'E:?`RZ^)FN@._
M>CK2W)!2LN9&1?=E'\OL_I4]S2-`Z6F)=MW3I.N>Q"XU42@..32.=V@61O1W
M!#(5FW@]/E0P)-QVL%49@>Y),,=-7Y\G/O=84N4I.&7I#(Y7TSMNK,I!-3'"
M:X3O(,0)%`^K)7Z2;BXXHHCG@?[4$C4BYQ%@HS@W@K<&+[2RI9;<SEY<0V:`
M8U9;UV%Z0#(3,)/!@!I\I-9H]37B;FL]\^&8=T(HT!7SF<?3.\@#1^'(3P(Q
M9A-L/:%I-=%@.[U'%>40I?X@T570N2-@351?_*;W<N<0[3#HE?H%KZW(6^AH
MHX:[=R/@V#O%G1L4__V#35(%H(PN"*<':E#:X_75C7*]7EZWGS3YEIHFF/RM
MI8ID+9[0@6*8M6W3O+;^PZR7L]#A;3$4':;;7K.[<N*N]%_<-0E#U)1/(^^R
M]4(+>)=)ER`OY2.8D^*"`9&"YN(DLU*4XM4RT"FM,H%RFFB1$E[X*;$6'Y7H
M_GT,"C)X_[?@I<KB_C2,DVZK=U34F$,?&$.HO@JS*9Z@+>+5`@8/82G\0[1]
MK+@X>W-VR$Y9?!VMIAB*UA2`_>75:@;Q*&;3:,J.]NJH%7C9$,+)-)P'/'#1
MA0!,2R!_D%NG"8]%H2&>IGA:XFGC%I&O`XXH<L73$\^FOLYF5\]2`->J[6A=
M-GK[-#2AT6O4W+U:^,RKUS.X+NDHSW<Y=V3<-JTMV)N2O7D_>Y;A3ZS;9FL+
M`9848#U"?^+<MIPM^-N2O_T(_L2Y;1M;\'<D?^<1_(ESV]YF?%W)WWT,?QI@
M9YL!]B1_[Q'\B7/;V69\FY)_<RO^.A@II^N;P)_BJO@E3*[UV8ZKQ+Z,,^R8
M6=CDX](?PL3UI]-;/#3VZ0R9`))5'$PAWX<)'ZVNKO?Y_GH<3I-@22$G#3:4
MBU`:<D]LVE6%&T_U0)VW!.X,_9BB,\::+Q&61/.`$<RHN.P=JZYT^$$H6G5C
M=SO,<V0F5[Q^56C'QP"7&KKZ(K8!F[@?J4M0Y\$P@CV)[,`@2,!<K`9*L\"/
M;SD8#&LYC@T(EL#+AEZ9#_=*:\F7"2.CEM[?C=IO884R,:+V4?:!=&^9FF=H
M&%-V>/R@'>1U@)*#$UIBTYTL/W9(M[+X7L_9P=(ZG*9NA1Z&NH545D="RVC)
M&C>]'L]R-GJ:1+N(#T?9+"W=*1V!3695[KG90T3.)9*9[<?IO2^GX=O3WW\Y
MJ0JP3F3?&&=XA:C^W^*CF.+QL^EP7BO&*:0E$(0PA'P0@XTKXSM7GNU,L"`/
M[X%U=WF4.V2OKP,$(.D$"78S7_`?_.\$9B@!MD!X_"O0:<).TE/E>IWI&]7N
M7H\?I>*DZG\XN^BSLW??B3U^)5B'2>WD]].+_MN7I^]__70B%P_DJ0ZB2EAJ
MASW?P'DC1%@4M)'T,5)ASI5?0Z[3/K;L<E(YSEF&[RA@LHR'`AU%$,JEX7H\
MRE5IF&WJESI]6BK/:\]AP\?,O595!"_SZ*;=T:[A']VH%*/DNNG-GEG>Z0Q<
M]1*$WOX+,P'MUAYO*J"G$J9JZR00?]A+A(32I/?<[.QDST5%/FMP#XE7U3JY
MOS0`(3TQD_EUD=LD,(&J<-]%,`!R=3,FT\BZIU%)DV@\?DB,N*RGM;A?1HX>
MMLX/2<`;AAKY_>PE,7XEP(N[:]VO\)U#4P(G,7,XB5B/Z&I=[GIK=]+KDO5[
MQ\=H';P6]Y:NHV%/^#*YL\,>T<Z00$'FI.I^Z1;G8F6D6P]++[8KD9ZB*!Q7
M.U)`#[R@\\(#MYOPR,$]`@:;:'"7`"LR<,D$L1(%C^2`&AT*Z>&X]L1Z(W&3
MRF/'IK/MV.2LTQ%6%8K>JR?]]8^`R,31WE9-1,?T&I6UE$$Z"DN&B$!'@2EF
M0W>LOK`IO](233]#4IE\)^?"!B"5I",OQJTJGGDL!>///L\K2JM44D8YAV"R
M\^U<Y"_DPT'!>QL(DU&>)U9#+;G5UP>9RZ;)S&YGP[[NIE'L0$_]?<Z;B#(4
M$',5).(8AD[=,5Y7L_SQ^"MW;_/#:C8(\*8SKA1QB!<G4V$$X1,JI]AD_AL%
M/!+X!&X5XKVX831;K!)^)``)@_5'[=D(TB:>$@EP#SJ,V]1ZK3:>1GY2GT97
M\F<J@;U@>O4H6H%6=6;M0XB@;JL_O9JD![BAT0C!$:U&:#="IQ&ZC=!KA$VZ
M%F-T0K,36IW0[H1.)W0[H=<)FYWTTK-!*Z'1SB)46M9KT&`1K4FT9CL+-VFT
M9DIK$:W5SB)'&JV5TMI$:[>S*)!&:Z>T#M$Z[2RBH]$Z*:U+M&X[B\YHM&Y*
MZQ&MU\XB+1JME](VB;;9SJ(F&FTS/5O-'ZZK/]B89/Y(8T(S3`<L"65!E!(1
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M;?-J/!WOF?O?$NC*V6.#9SE7;G'5;+'%QR_/`G2)2S<*T>4.W2A87^'2!>L;
M.W1EJ+XD=NE"]26I0U<$Z\M<NF!]I4.7!^L;N72A^E*W/[*79WBWOYTX<YGD
MH4HSMU-X%J)SF8R#];F=$H?JRYU!$[\)U9>G+EVP/F?0Q*^#]94N7:B^PADT
M\:M0?05WZ8+U.?T1OVRKKYK>#JLC\?]C^'^]ZU+@^PL[3IB-N^2XTCZ^PB4X
MO;&P,W#[)F9?LCY;BH_0P_A`O^B3@_GD`(]#XLD$;[>)!F'[A"AY&R47E`R!
M$.%03S`GZ&5+/P*_N[+AI-$P3[SZQ,/8JGPB;SSA/X`,)&VLIFV4*5"V<)IK
M3EUQ_4J(OY8*$E?4`AS`CV^;`H]6"SP./)>U"9.Y8B-AWD:8"T)@>>#R//:%
M)A;]MI#%EL<+1^0DMIBFBL8M%6E1D]2G+]I8+R:31L^6;82EE)&Y7(IFVGO6
M5Y26TJ\@=L7,`CV4^"^1V[-)&7C.'Q`HS*AM0/OJ1,IQ0.PR(+:OY*#8F2OV
MJ"FP;PM<@=.FF>&M=H;':#X:G+:;&MXN;\H#\OIC)2COR%1P;%60A"R4(`-,
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MPY.DWQ#_0<5Z\$#?W4V?QBY0``3^GD^YV3X6??LS$9[)R*`1U]/3"OWJU_X!
M?^ZPY]=L#\]Q_OI^OA;U>KBQL?8]WM:2!>C8.F>C+01Z.:MNI\?547TIWNME
MM5A6U]5<'DQ>$5``.C#"79A9Q>2A1W5LU_[Z=GJYN*#&Q3=>FK"J9,<IBT_8
M"6?)F,U200Y0E*Y(]1RN\^,9.&YQO#SS","F\,0B249GS*,A&VD3I=D978G[
M(/($BSI_=(;9<\C6F2\IL[8SLY@RM0G<M<HR*KMT'I!5G^^*YDRNJ%M?<F8[
M[/ME+10^G5U]E+@55_-/U?*&S3Z>LMT!\`R>"`R'DBP[G"Y/KZ4G]H;X^Y/Q
M),"K"]"<;"V/82,DZI'5K=G>/BOEF2(^?[1+ESR/P$*+F@[X9CT1RVF9#5=9
MX9GU\;HY4CMB6R(G7D>(SO@VA5D?P#]/*X7W!8>'T^NCNF8WT]E%13=^Z')C
M_VAO+U-^SX[K-%;;7Y^NLRW&X\%NRU,]^5B#8?#?:^/WWV3WQWU2T/U9Q2<4
MFS5>9F@Z=^.M3/A#"'`T9(V./.EC"=SI&4-G]<RU-GRWAR1Y/-%_(`O6\9<"
M''E37US(UY61ER44.,,(FU57(>!^-[8$&!\[`]O]\_T<#<4.OMSPP+F\)D-#
MZWQO7PTL?<P74_0(H1D)DGBN%!)&@+ET:KW$90)5VP+\TNM=RAKO`P&C7UQL
MHJ?:,-E[C.?JP+V!_L(!]/"^@"^F+4?B._%?''W3.?J.TO>'?:.7#_:+[.N[
M!1;F`U[[Z'TPW+3[!_^Y#SSM]=_UT;)>W%P_>AS(^\9_S,LBYN)OB/_'N_B/
M3Y+:^A\NYVXO/C]:&W?T?Y8F,:[_.<]2BO^8E7G6K?^?(GWQEQ<?KY<O9O7\
M137_Q!:?;\[$I!-]P18(BT'KV[=?1X?U0DP1!U$3]]Z`W"?#J`EH/]3H]9DH
M;B#5#S4L?2&*LS9X?L(C&@TC`'D">/E,@0_E!#<$F.ZB=$PP\@@*GQ,T$9=P
M[ZFHFQ/Z$^(1E83&E,C@#KDHAG]R@DP"N"*`3TIDE(PRFHCEZ:%X3:9B17TU
M/\0E$:HC5>$@5,B)3`;MR*R`!2JH"@1UD:%+5-`6+@,BR']U`!A.`%:1_)=4
M/);834)8^2^A5J6D):&.*%&X6`4I`/4OVI7_HL(1,1_Y!KD.3ZAC0=\C`[T%
MW0"509<(X@AT4I)Z@#]0%;1;$`)_!)U94+^""-#'P!KTMV@R0FA_TCWJ.R7N
MH4^`.\2)D@A;B83CBBDZ`D!R(9P5=2SH##H9=)%CE`%1#(W1B`2UPN@$=>4$
M;Q7!2.$T:,8T?D"A.08```60^&H5`[*71L\J.Y&!H4"/O#3Y,J8+=D%ALG-B
M(>56<!*&@V0D1W1F<G&4ERCBV*I9AL*`'K-J3F34#>QGS!4BP#WJ>O[I<%:?
M]]\-=C#[K7B;A6"WB*!Q"XO=V^5T?EKU+ZJY(!I,D`H*ZV;ACFX/",Y7$>#)
M\`E[)]:S<*6\=DLD)_3%<MXH0H`6S)6++4%KQ"$3A!+UI#@3O"?:)H^@`:+M
MZ6)1B17>N^UZ?ES=]F\W$:G`JKX740N'RL)5?7R?APRO7D)%:VMK7ZM"L6H&
M9T^\E8E>M["*EA#,@``(CPIZ:.+ZXP4@3XAO#[;!@"TL%*W7^W'4^X(^(K`F
MN685JM^S"$7+H`I)6-,OJI84B,\>(+'(F&QQ(JG9YKYXF0V@#3Q!0F)5Y*7;
MQ_]"(PA#4,]/]]?7)6^@4X"A@(\^H-K![T;X=-X7PQ-^R4=$0^OOY^NJ^#D;
MH8>N10&\K!^LP]+:9*P]/QZN"6KIOE)O[O.[JI@,UZ/(K@(W4D@UE$L2DCO?
M#MQH3/`K5H^;A="\CL$ABN1X*3)]VUPPNCEA/[#]=9DA-:6?PGW<>B'=G+WQ
MIZE06Y)(C3\F)LR#6]%%$+S1;NWMU_T?!HT6U>.#79?Z71PD/=A)$SH`?X>?
M[W:VQ>4[U&R-)Q#O8O56#"@/-+9(9"8"P?=4Z^\!!XXA/OL/?__F]2L:_#H/
M_(PD+`_<R%\7_=B_%=9JJX_5/F?90!BJ38:_7H"Y@K\\77RS0CHQ[%`Z[0@?
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ML++A'DIL#/JHA$W![P:PN?GL<@/.0N/-9_7FL_,-51,UO;D)/\0TXG"`G$O6
MD0()7/E^'5H"7@(5$BF"7W=%L5(QL+$90;G;:R<GT3\_UC>=*?UO2$'[?_*$
M]K],T/XG<5ZF28+V/R\Z^_\4Z='L_\G=]C\>\7R\:@)(TGM-`"?WG0!&GOUG
M[&<`,+BZW8%UZR[[B@)XZ&?%?R]_T^R`!T"//3OP\I&G!P@KVDT/7;)3T/XO
MG\[^YTFNUO\EG?\4,"5T]O\)TJ/9_^5][+^8UO-[VG\K,$EH'EBJ>2"?M,\%
M0V9F`SNVR2#2.Y'6U`">,B`%S0W?X!$TP(@Z\X,YQ.8]94CGSE0QQ^\.9ZM3
M3RAVICUWZ+9WK,;EWU\KK)T@(^+7W*G:GHFLF<N:0FCZ<N8O.8-A?MLTYL]C
M[%X3V5Q/9)IAR.%98VYS)C<UNSE<RBG.GN-L.EL8-=')#D):.25&-J4S(5H7
M%**VF=&9&'M_CJG1MO_5_&9YM?C\Q/<_>9:*,CC_+Y.DH//?M.#=^>^3I'_C
M_J?)NYS>G#DY:[0;M!8UW(M-?$H?U*$W<HH,H$+/*Y%@-W:V!XC?Z_'$-"2Q
M5=FWWWTS9&]?_<`&K,_ZXM?@H"]^#EZ,)NR7?=87$Q2'>^?]DFVQ/A8]'PT&
M`]C!]ZZ8PN+H($EQMA'6UK^!.B"OG&&HA`=+DF!)&BS)@B5YL*0(EI3!DE&P
M9!R6=(42PEK@837PL!YX6!$\K`D>5@4/ZX*'E<'#VDC"VDA6C`G0!J"#^U'H
M8!#R+#@(Z^`@K(.#L`X.PCHX".O@(*R#@[`.#L(Z.`CKX""L@X.P#@_".CP(
MZ_`@K/4@U`$P`C?@VX,$ROB8@^AG%O*C3W>UTW;"0Q[\A4T4GP6J`OP#=6G\
M=1#\8&01Q2'$`&Z(7F4(`Q#=Z4X^K*Y;7<H7!QP@MK!D]J]J>=5?#"6PT$+>
M*`7_13C5)/?&R\E!/0>/1E7(=2&7A=P4)KHPD86)*4QU82H+4U.8Z<),%F:F
M,->%N2S,36&A"PM96)C"4A>6LK"<*/G)TZ@O':-`[`'[44Q0**-R+05"'!K&
MA8K>!J@./(X"E26ZLM2MC-R2Z+U1=4@_=%,/M2B]GQ0+A<<"3^[@(=,\Y$$>
M>+R2B<)F@CQE'1;2.U@H-`MEF`4>9`%:=!@P#V6F92RZKJ6[WY>6%Z_D"L(2
M0=STY7+*%AA@CEU_!$>'?WZL\$E%5<]/+D36<;T$^"L(QB`(R<GC<N))B+YH
M,%900NU>AZ\/^=!9P\=%0'!'D'9;;*L^;52?>-5[OG'>R!KX/6H@$C1OHP9O
M9F@9A\(V[O(&=]DJ[KP1UV!MY+.6V,[GWHA;J;:RP5AQ#[59([&AMX0[(['M
MZ<Q^VB*3/HC*(M-H8A+*=*#KM".B*-!>&>U$1@F8$VHOQL5B&]_*`$6?;RKZ
MZZP^/9/`M2T1<3?@NAU$M5((BRK$Z+=L8/PUMCC>@Q&DW^XZ84#\,`M.8\H9
M!\!3X06`$.@QPL"*GWV^MU</9%X+?AS(OO57NN""&RVC#9!IL]R2<0][J`B3
M._'PU[V0AYJ#%@9^C_9_=1T6L*].Z\O+ZE!^UIO^0AA,MB'S[^@DMP1#^N!&
M&_0+UK-87LVF*DN"I9M._>47]A?=4$O_-A98T.&T)L"VY+J`?J#'AEPP0*NJ
M$'\,!@IY<[<F_$,S4HAM`.6?&'</$QL*J)UA1?3U1#M,D9`$12_+V`M`/MV.
M=Y7^HYY2*;R46&!')C`<-5O0.W<]4\?6/M.M;BCL4MAKZNL?JESYCEGMJS]?
MV$\R%_?4\W&Y^;RHX$:=VR,(BKH;V<'>9&PZ.526XBO[5@<F$V5D!ISX[V9(
M$+F^9P8N.M=L;2BS=0\*A4$%6F,F'!OY\XAFI-SAJ'<J/!)!F0,P/?&(]_9:
M`BB#-Y*^TZ?18&].XV8$;R!5R_Z!';E)MD7_Q+NMV9RLJ:I_>EK%*K:?*&@)
M%TZJ;P?`=X,30BAQ$RE4<F'C=LHL[=SV5CJ;_:M21=&#(H[+A[91;T/U"Y1F
MHI!;;<#=V2)C4@@5*]&P+!74Y!C]XMNC;LCM':7;>B*9H4`Z*T(PWD-,3TZK
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M7TX';(_%`SJUP5`>%%:B-X,;W%.:I9P.%+P>/#^>,%S(84@0`/S&2'57)SNP
M#*(>;>6I9]:B+B/L?$@-M[=U<?738S858X?*=S_99M]!'/(C*_Z+T.V-&/!S
M#8J.\2P%`W#O6C0_+KQ>_8[H9Q;Q_!-`W@.:-*T.$?!\IH92NLW>@'OKQ>>A
M]%6&9DV#-_5EQ::XN*PQ^M>Q#]@^TW37+B&-(7NU)%[&>G-_9-FAIKJ,U@_H
M]9S`828X/F^>#^F30OQ%"Q?OG9.*S+;9WUY_CX@`T:KH+TVCU%C8_-%G+/^?
MDX?_@D[23XS_6W`>&_Q?GN#Y7UIVYW]/D1YZ_K=VL@HF6%T:!3`/L8BNQ!O8
M#E3"S.XU()4\"-]%[Z)_".((B^6"0ON(OPKBH^[VF.$C_M_0_OC(HPO5QUW\
MUC!X\6A7,[<*0=@0A5I,$D.T"CO8$&5GO0!58:CREX(JL+D_\S;W:8-I1AB'
M)[1;_V%R`/V,D*^T$1K;^U62CBLZ;J&HFOU5J-/;AS/PAZH*C&TCLJR&7KS@
MB<64WGL[H?,`V:;$>&TPE2J"]#Y,N?*DFAF"-Q4C@:?6PR1"RW:Z!P]W(F$%
MM6@@TZA%TYEB-ENMZ5S1Y0\7*K,TC`^"5)DG5>%+-0I)-7:E&D?!KBH4TT6@
MJTI%4#Y<JMR72O"2>T(UMH%Y"Y3?B80!-%()_21Q4*J18GH4D&JL",:3R0.%
M*EJ$*CRA>!P\2E'"<$\8'NXCKM]U'@?DX>8UO]=[[CY<>J^4D*BTG_4.9JYK
MUI3'LA&NN6I:*]J:;WF1S!Y[B^FZPVIQ\S+XZC.LX2D"(^-E\;C*?,E'PERZ
MMNQ.G?LRINXX8`U;UG)0$@=%'3FBMM@T[P''MMTIJFOA'BQJYEHEV[S1J8F<
MD8V@25#0L=NG#2NWRLRIPZ0@IZ[->["<N2NG;_&2I-&A64A.R^;I#A5V+RRH
M8_GNDM.V@K]!3ML2XH3EF\&D,0\G15!2[K^E#[.*#5D]9AT#^7!92[=/2U_2
MXLXJ'`L)LB7A4T/KB-`^..R^L?]3D_?]#QNP3X[_FG.NO__CO,-_?<K4N++;
M_F&OT`"BWPLOMKLMU]V6DX7=;;GNMEQW6ZZ[+?>?=5ONCY[(N]2E+G6I2UWJ
M4I>ZU*4N=:E+7>I2E[K4I2YUJ4M=ZE*7NM2E+G6I2UWJ4I>ZU*4N=:E+7>K2
,?TWZ/VX96&``\```
`
end
--[ EOF
Reference in a new issue Copy permalink