feat: finish xiaosai

2023-10-15 19:42:16 +08:00
parent 7fe37b52eb
commit c2de294d1c
13 changed files with 571 additions and 0 deletions
--- a/xiaosai/Claude_Shannon/crack.py
+++ b/xiaosai/Claude_Shannon/crack.py
@@ -0,0 +1,46 @@
+from pwn import * # type: ignore
+# context.log_level = 'debug'
+
+"""
+(这题小小的放了水)
+因为只有一轮,而最终的答案有128种情况,那我们不妨直接猜一个答案
+根据条件概率公式算一下
+
+for i in range(0,1000,100):
+    print(i,":", 1 - (127/128) ** i)
+
+0 : 0.0
+100 : 0.5435690025908804
+200 : 0.7916707446041162
+300 : 0.9049120701701576
+400 : 0.9565989213461966
+500 : 0.9801904023814129
+600 : 0.990958285600675
+700 : 0.9958730812784277
+800 : 0.9981163463716863
+900 : 0.9991402420956556
+大概300次以内都出结果了
+"""
+
+
+
+for i in range(300):
+    try:
+        r = remote("localhost", 10011)
+        # r = process(['python3', 'main.py'])
+        for i in range(15):
+            r.sendlineafter(b":", b"1")
+            r.recvuntil(b"!")
+
+        r.sendlineafter(b"chests:", b"0 0 1 0 0 1 0")
+        r.recvline()
+        flag = r.recvline()
+        if b"flag" in flag:
+            end_time = time.time()
+            print(flag)
+            break
+        else:
+            r.close()
+        
+    except EOFError:
+        pass
--- a/xiaosai/Claude_Shannon/crack.sage
+++ b/xiaosai/Claude_Shannon/crack.sage
@@ -0,0 +1,195 @@
+"""
+前置知识（建议要看，非常自信可以不看）
+https://zhuanlan.zhihu.com/p/448198789
+
+信息论的一些基础
+可以结合这份ppt来看
+https://web.xidian.edu.cn/jwliu/files/20190521_003040.pdf
+
+同时b站上的国防科技大学-信息论与编码基础也可以学习一下
+https://www.bilibili.com/video/BV1pJ411U7G8?p=61
+
+
+读题可知，我们需要获取一个7bit的有效信息，每次返回的信息可以看作1bit，共15bit。
+而其中1-2次的说谎我们可以视为1-2bit的传输错误。
+因此我们需要一种最少可以纠正2bit的编码，也就是循环冗余校验
+
+根据题目描述，我们需要构造一个（15，7）循环码，也就是15bit的编码数，7bit的有效信息数
+"""
+# 首先定义伽罗瓦域
+R.<x> = PolynomialRing(GF(2))
+
+"""
+然后寻找生成元g
+根据定理，（n，k）循环码的生成元g一定是x^n+1的n-k次因子
+（ppt中说是x^n-1,这里因为是GF(2)，所以+1和-1是等价的）
+带入计算得知，g应当为x^15+1的8次因子
+"""
+ff = x^15+1
+g = factor(ff)
+# (x + 1) * (x^2 + x + 1) * (x^4 + x + 1) * (x^4 + x^3 + 1) * (x^4 + x^3 + x^2 + x + 1)
+factors = []
+for i in range(len(g)):
+    factors.append(g[i][0])
+
+possible_g = []
+
+# x + 2 * y + 4 * z == 8
+# root = [(8,0,0),(6,1,0),(4,2,0),(4,0,1),(2,3,0),(2,1,1),(0,4,0),(0,2,1),(0,0,2)]
+
+
+possible_g.append(factors[0]^8)
+
+possible_g.append(factors[0]^6 * factors[1])
+
+possible_g.append(factors[0]^4 * factors[1]^2)
+
+possible_g.append(factors[0]^4 * factors[2])
+possible_g.append(factors[0]^4 * factors[3])
+possible_g.append(factors[0]^4 * factors[4])
+
+possible_g.append(factors[0]^2 * factors[1]^3)
+
+possible_g.append(factors[0]^2 * factors[1] * factors[2])
+possible_g.append(factors[0]^2 * factors[1] * factors[3])
+possible_g.append(factors[0]^2 * factors[1] * factors[4])
+
+possible_g.append(factors[1]^4)
+
+possible_g.append(factors[1]^2 * factors[2])
+possible_g.append(factors[1]^2 * factors[3])
+possible_g.append(factors[1]^2 * factors[4])
+
+possible_g.append(factors[2] * factors[2])
+possible_g.append(factors[2] * factors[3])
+possible_g.append(factors[2] * factors[4])
+possible_g.append(factors[3] * factors[3])
+possible_g.append(factors[3] * factors[4])
+possible_g.append(factors[4] * factors[4])
+
+"""
+至此我们已经找出所有的生成元了，但是不要忘了还有纠错的要求。
+纠正2bit的要求则是： 假设用d表示码组的最小汉明距离，纠正错误时，设可纠正t位的错误，则d >= 2t+1
+（也就是保证任一点A的错误状态不落到其余点B，并且A的错误状态离A更近）
+带入公式则是d >= 5
+"""
+
+# 定义函数 n2p，将二进制数转换为多项式的形式
+def n2p(a):
+    a = bin(a)[2:]
+    p = 0
+    for i in range(len(a)):
+        if a[len(a) - i - 1] == '1':
+            p += x^i
+    return p
+
+# 定义函数enc，将7bit的信息编码成循环码
+def enc(i):
+    m = n2p(i)
+    c = (x^8) * m + (((x^8) * m) % g)
+    return "".join([str(int(i in c.exponents())) for i in range(15)])[::-1]
+
+# 定义函数dif，计算ab之间的汉明距离
+def dif(a, b):
+        cnt = 0
+        for i in range(len(a)):
+            if a[i] != b[i]:
+                cnt += 1
+        return cnt
+
+"""
+通过构造校验表来检查所有的生成元是否满足条件
+"""
+for g in possible_g: 
+    # 创建一个列表 dic，存储不同输入值的编码
+    dic = []
+
+    for i in range(2**7): 
+        dic.append(enc(i))    
+
+    minHD = 15
+    for i in dic:
+        for j in dic:
+            if i == j:
+                continue
+            if dif(i,j) <= minHD:
+                minHD = dif(i,j)
+
+    #if minHD >= 5:
+        #print(g)
+    # x^8 + x^7 + x^6 + x^4 + 1
+    # x^8 + x^4 + x^2 + x + 1
+
+
+# 下面两个随便挑一个作为生成元就好了
+# g = x^8 + x^7 + x^6 + x^4 + 1
+g = x^8 + x^4 + x^2 + x + 1
+dic = []
+
+for i in range(2^7): 
+    dic.append(enc(i))
+
+"""
+构造问题这个地方有点巧妙。
+我们可以发现系统码是一个对称互补的码，也就是说0和127，1和126的系统码每一位都相反。
+
+那么对于所有系统码来说0和1就是平均分布的。（因为成对来看，30个bit中必有15bit的0和15bit的1）
+对于系统码的任意一位为0的概率就是0.5，为1的概率也是0.5
+
+因此我们可以把questions[i]和j[i]等价来看，
+我们可以找出所有令j[i]==1的信息码情况，共64种，把这些进行or运算。
+把这个问题发送出去，如果返回true，那么证明j[i]就是1。
+以此类推，我们就可以得到一串15bit的返回值。
+
+但是因为存在最多2bit的错误，我们需要跟码表进行对比，找出汉明距离小于等于2的系统码。
+然后这个系统码对应的信息码就是正确的信息码
+"""
+
+# 定义字符串模板 part，用于后面构造问题
+part = "( C0 == {} and C1 == {} and C2 == {} and C3 == {} and C4 == {} and C5 == {} and C6 == {} ) "
+
+# 创建问题列表 questions，用于存储构造的问题模板
+questions = []
+for i in range(15):
+    question = ""
+    nums = [j[:7] for j in dic if j[i] == '1']
+    for j in range(64):
+        n = nums[j]
+        question += part.format(n[0], n[1], n[2], n[3], n[4], n[5], n[6]) + "or "
+    questions.append(question[:-4])
+
+
+
+# 导入 pwn 模块，用于与远程或本地进程进行交互
+from pwn import *
+
+# 设置日志级别为 debug，以便在执行过程中输出详细日志信息
+#context.log_level = 'debug'
+
+r = process(['python3', 'main_local.py'])
+#r = remote("localhost", "10011")
+
+for i in range(1):
+    msg = ''
+    # 逐个发送问题并接收答案
+    for j in range(15):
+        r.sendlineafter(b"Shannon:", questions[j].encode())
+        r.recvuntil(b"answers:")
+        if b'True' in r.recvuntil(b"!"):
+            msg += '1'
+        else:
+            msg += '0'
+    # 查找与收到的答案最接近的编码，这里答案肯定只会有一个
+    for j in range(128):
+        if (dif(msg, dic[j]) <= 2):
+            break
+    ans = ""
+    for k in dic[j][:7]:
+        ans += k + " "
+    # 发送解码后的答案
+    r.sendlineafter(b"chests:", ans.encode())
+
+flag = r.recvall()
+print(flag)
+r.close()
+
--- a/xiaosai/Claude_Shannon/main_local.py
+++ b/xiaosai/Claude_Shannon/main_local.py
@@ -0,0 +1,57 @@
+from random import choice as c
+from random import randint, shuffle
+from Crypto.Util.number import * # type: ignore
+
+flag = b"flag{this_is_a_test_flag}"
+
+# 白名单列表，包含允许在问题中使用的关键词和符号
+white_list = ['==', '(', ')', 'C0', 'C1', 'C2', 'C3', 'C4', 'C5', 'C6', '0', '1', 'and', 'or', 'not']
+
+# 定义一个函数 calc，用于计算问题的结果
+def calc(ans, chests, expr):
+    try:
+        C0, C1, C2, C3, C4, C5, C6 = chests
+        r = eval(expr)
+    except Exception as e:
+        print("Shannon fails to understand your words.\n", e)
+        exit(0)
+    return ans(r)
+
+# 定义一个游戏回合函数 do_round
+def do_round():
+    # 定义两个函数 truth 和 lie，用于判断 Shannon 的回答是否为真或假
+    truth = lambda r: not not r
+    lie = lambda r: not r
+    chests = []
+    for i in range(7):
+        # 随机生成 7 个 True 或 False，表示每个箱子是宝藏还是怪物
+        chests.append(c((True, False)))
+    print("Seven chests lie here, with mimics or treasure hidden inside.\nBut don't worry. Trusty Shannon knows what to do.")
+    # 随机选择 Shannon 的回答策略，包括多少次 lie
+    lie_count = c((1, 2))
+    Shannon = [truth] * (15 - lie_count) + [lie] * lie_count
+    # 猜猜第几次是在说谎 :-)
+    shuffle(Shannon)
+    for i in range(15):
+        print("Ask Shannon:")
+        question = input().strip()  # 输入问题
+        for word in question.split(" "):
+            if word not in white_list:  # 检查问题中是否包含白名单以外的关键词或符号
+                print("({}) No treasure for dirty hacker!".format(word))
+                exit(0)
+        res = str(calc(Shannon[i], chests, question))  # 计算问题的结果
+        print('Shannon answers: {}!\n'.format(res))
+    print("Now open the chests:")
+    # 输入0或1表示每个宝箱的真假，用空格隔开
+    chests_string = input()
+    return chests == list(map(int, chests_string.strip().split(" ")))
+
+# 游戏开始
+print("The Unbreakable Shannon has returned, with some suspicious chests and a far more complicated strategy -- he MAY LIE ONCE OR TWICE! Can you still get all the treasure without losing your head?")
+
+for i in range(50):
+    if not do_round():  # 执行游戏回合
+        print("A chest suddenly comes alive and BITE YOUR HEAD OFF.\n")
+        exit(0)
+print("You've found all the treasure! {}\n".format(flag))  # 赢得游戏，获得flag
+
--- a/xiaosai/Claude_Shannon/main_remote.py
+++ b/xiaosai/Claude_Shannon/main_remote.py
@@ -0,0 +1,100 @@
+from random import choice as c
+from random import randint, shuffle
+import socketserver
+import signal
+import string
+import random
+import os
+
+class Task(socketserver.BaseRequestHandler):
+    def _recvall(self):
+        BUFF_SIZE = 2048
+        data = b''
+        while True:
+            part = self.request.recv(BUFF_SIZE)
+            data += part
+            if len(part) < BUFF_SIZE:
+                break
+        return data.strip()
+
+    def send(self, msg, newline=True):
+        try:
+            if newline:
+                msg += b'\n'
+            self.request.sendall(msg)
+        except:
+            pass
+
+    def recv(self, prompt=b'[-] '):
+        self.send(prompt, newline=False)
+        return self._recvall()
+    
+    # 定义一个函数 calc，用于计算问题的结果
+    def calc(self, ans, chests, expr):
+        try:
+            C0, C1, C2, C3, C4, C5, C6 = chests
+            r = eval(expr)
+        except Exception as e:
+            self.send(b"Shannon fails to understand your words.\n")
+            self.send(e)
+            exit(0)
+        return ans(r)
+    
+    # 定义一个游戏回合函数 do_round
+    def do_round(self):
+        # 定义两个函数 truth 和 lie，用于判断 Shannon 的回答是否为真或假
+        truth = lambda r: not not r
+        lie = lambda r: not r
+        chests = []
+        for i in range(7):
+            # 随机生成 7 个 True 或 False，表示每个箱子是宝藏还是怪物
+            chests.append(c((True, False)))
+        self.send(b"Seven chests lie here, with mimics or treasure hidden inside.\nBut don't worry. Trusty Shannon knows what to do.")
+        # 随机选择 Shannon 的回答策略，包括多少次 lie
+        lie_count = c((1, 2))
+        Shannon = [truth] * (15 - lie_count) + [lie] * lie_count
+        # 猜猜第几次是在说谎 :-)
+        shuffle(Shannon)
+        for i in range(15):
+            self.send(b"Ask Shannon:")
+            question = self.recv().decode().strip()  # 输入问题
+            for word in question.split(" "):
+                if word not in white_list:  # 检查问题中是否包含白名单以外的关键词或符号
+                    self.send("({}) No treasure for dirty hacker!".format(word).encode())
+                    exit(0)
+            res = str(self.calc(Shannon[i], chests, question))  # 计算问题的结果
+            self.send('Shannon answers: {}!\n'.format(res).encode())
+        self.send(b"Now open the chests:")
+        # 输入0或1表示每个宝箱的真假，用空格隔开
+        chests_string = self.recv().decode()
+        return chests == list(map(int, chests_string.strip().split(" ")))
+
+    def handle(self):
+        signal.alarm(60)
+
+        # 游戏开始
+        self.send(b"The Unbreakable Shannon has returned, with some suspicious chests and a far more complicated strategy -- he MAY LIE ONCE OR TWICE! Can you still get all the treasure without losing your head?")
+        for i in range(50):
+            if not self.do_round():  # 执行游戏回合
+                self.send(b"A chest suddenly comes alive and BITE YOUR HEAD OFF.\n")
+                return
+        self.send("You've found all the treasure! {}\n".format(flag).encode())  # 赢得游戏，获得flag
+
+
+class ThreadedServer(socketserver.ThreadingMixIn, socketserver.TCPServer):
+    pass
+
+
+class ForkedServer(socketserver.ForkingMixIn, socketserver.TCPServer):
+    pass
+
+
+if __name__ == "__main__":
+    flag = bytes(os.getenv("FLAG"),"utf-8") # type: ignore
+    # 白名单列表，包含允许在问题中使用的关键词和符号
+    white_list = ['==', '(', ')', 'C0', 'C1', 'C2', 'C3', 'C4', 'C5', 'C6', '0', '1', 'and', 'or', 'not']
+    HOST, PORT = '0.0.0.0', 10011
+    server = ForkedServer((HOST, PORT), Task)
+    server.allow_reuse_address = True
+    print(HOST, PORT)
+    server.serve_forever()
--- a/xiaosai/ecb_padding/crack.py
+++ b/xiaosai/ecb_padding/crack.py
@@ -0,0 +1,37 @@
+from pwn import *
+from flag import flag
+from Crypto.Cipher import AES
+from hashlib import sha256
+import base64
+
+flag = b''
+first_flag = b''
+wordlist = b'123456789qwertyuiopasdfghjklzxcvbnmQWERTYUIOPASDFGHJKLZXCVBNM{}_'
+for i in range(16):
+    for j in wordlist:
+        r = remote("127.0.0.1", 10002)
+        payload = b'0'*(16-5) + b'0' * (15 - i)\
+            + first_flag + j.to_bytes() + \
+            b'0' * (15 - i)
+        r.sendline(payload)
+        cipher = r.recvline()
+        cipher = base64.b64decode(cipher)
+        if cipher[16:32] == cipher[32:48]:
+            first_flag = first_flag + j.to_bytes()
+            break
+        
+last_flag = b''
+for i in range(21-16):
+    for j in wordlist:
+        r = remote("127.0.0.1", 10002)
+        payload = b'0' * 11 +  j.to_bytes() + \
+            last_flag + b'0' * 27
+        r.sendline(payload)
+        cipher = r.recvline()
+        cipher= base64.b64decode(cipher)
+        if cipher[16:32] == cipher[64:80]:
+            last_flag = j.to_bytes() + last_flag
+            break
+                
+
+print(first_flag + last_flag)
--- a/xiaosai/ecb_padding/flag.py
+++ b/xiaosai/ecb_padding/flag.py
@@ -0,0 +1 @@
+flag = b'flag{eCb_is_not_SafE}'
--- a/xiaosai/ecb_padding/main.py
+++ b/xiaosai/ecb_padding/main.py
@@ -0,0 +1,73 @@
+from hashlib import sha256
+import socketserver
+import signal
+from flag import flag
+from Crypto.Cipher import AES
+import base64
+
+
+
+
+class Task(socketserver.BaseRequestHandler):
+    def _recvall(self):
+        BUFF_SIZE = 2048
+        data = b''
+        while True:
+            part = self.request.recv(BUFF_SIZE)
+            data += part
+            if len(part) < BUFF_SIZE:
+                break
+        return data.strip()
+
+    def send(self, msg, newline=True):
+        try:
+            if newline:
+                msg += b'\n'
+            self.request.sendall(msg)
+        except:
+            pass
+
+    def recv(self, prompt=b'[-] '):
+        self.send(prompt, newline=False)
+        return self._recvall()
+
+    def task(self):
+        key = sha256(flag).digest()
+        key = key[:16]
+        aes = AES.new(key, AES.MODE_ECB)
+        data = self.recv(prompt=b'')
+        data = b'cqupt'+ data + flag 
+        # len(flag) == 21
+        # wordlist = b'123456789qwertyuiopasdfghjklzxcvbnmQWERTYUIOPASDFGHJKLZXCVBNM{}_'
+        if len(data)%16 != 0:
+            pad = b'0' * (16 - len(data)%16) 
+            data = data + pad
+        cipher = aes.encrypt(data)
+        print(aes.decrypt(cipher))
+        cipher = base64.b64encode(cipher)
+        self.send(cipher)
+        
+
+    
+    def handle(self):
+        signal.alarm(60)
+        self.task()
+            
+        return
+
+
+class ThreadedServer(socketserver.ThreadingMixIn, socketserver.TCPServer):
+    pass
+
+
+class ForkedServer(socketserver.ForkingMixIn, socketserver.TCPServer):
+    pass
+
+
+if __name__ == "__main__":
+    #flag = bytes(os.getenv("FLAG"),"utf-8")
+    HOST, PORT = '0.0.0.0', 10002
+    server = ForkedServer((HOST, PORT), Task)
+    server.allow_reuse_address = True
+    print(HOST, PORT)
+    server.serve_forever()
--- a/xiaosai/factor_2048bits/crack.py
+++ b/xiaosai/factor_2048bits/crack.py
@@ -0,0 +1,11 @@
+from Crypto.Util.number import *
+
+p = 170021659067442061523756541191993539248601165314725269340063885026090819928194021233489114706002869522375199940625136026822330472394573193982550217044532146962203411567638298486561460035048451998397915884861905155825573014911356931131393751111816934458614898785910805905358218404703141056982153504282962094549
+q = 156289115230564684236869282390677373336412812898135298853785374595122098483952732832928286906302804608150589957073534308140034369315958525768786913022949461790625449368299077456798653080889386800523502656112132648850626018430482876211766660346216029055456596647573751654709671998718949348538504201995353414849
+n = p * q
+cipher =  15315994859113745140937150782850225701835894977924847828178261709106790713756481438851906169833437205680746393379892960895453635059199072504132009117455118252801524117640663250753749247480223306967637814097656959416506419849108127590869410005001001319682678605925584319221815990695423909010895701674831442387195906671891136861712786777889185580869108974742614564867198257449323144696618681193168712805318310217437715668293800077584570491524233826384775049353844578873853868493454822838612576965182892687235411858595304846022012252673457313575555164210208501163855569260816910756715565420825070869972114502924134935227
+phi = (p - 1) * (q - 1)
+e = 65537
+d = inverse(e, phi)
+plaintext = pow(cipher, d, n)
+print(long2str(plaintext))
--- a/xiaosai/factor_2048bits/flag.py
+++ b/xiaosai/factor_2048bits/flag.py
@@ -0,0 +1 @@
+flag = b'flag{W0W_you_cAn_factor_2048bits_prime}'
--- a/xiaosai/factor_2048bits/main.py
+++ b/xiaosai/factor_2048bits/main.py
@@ -0,0 +1,19 @@
+from flag import flag
+from Crypto.Util.number import *
+
+# Is it possible to factor 2048 bits RSA????
+
+p = getStrongPrime(1024)
+q = getStrongPrime(1024)
+n = p * q
+e = 65537
+message = bytes_to_long(flag)
+cipher = pow(message, e, n)
+print(p)
+print(q)
+print('n = ', n)
+print('cipher = ', cipher)
+'''
+n =  26572534665683235245636883276068606612246011683144885473463339711820331993493357279285848391008791322234296406608539840620252625886064149394162998948970494176467317183311068211754639508369506106626167584042532674048767021259563143175030276397301094824563794971045331643148544039656526615133318494601670688100830427083371422333448416345157657081894568628680132969792458721329931299527818220689347292610785587787837487603664368727957712489422350366840751269348100479824826707549651813948243868760028344290958809882314383956517334878930614926458454637703779278161529934602431923391507121553686269628353968327172758558101
+cipher =  15315994859113745140937150782850225701835894977924847828178261709106790713756481438851906169833437205680746393379892960895453635059199072504132009117455118252801524117640663250753749247480223306967637814097656959416506419849108127590869410005001001319682678605925584319221815990695423909010895701674831442387195906671891136861712786777889185580869108974742614564867198257449323144696618681193168712805318310217437715668293800077584570491524233826384775049353844578873853868493454822838612576965182892687235411858595304846022012252673457313575555164210208501163855569260816910756715565420825070869972114502924134935227
+'''
--- a/xiaosai/factor_n/crack.py
+++ b/xiaosai/factor_n/crack.py
@@ -0,0 +1,14 @@
+import gmpy2
+from sympy import nextprime
+from Crypto.Util.number import *
+n = 119089874014927074688341672884508166875005676500322264758691717600758737133958853429775347698400246542956663932584349274109281162512900740554148968627493442686180696496512753102593837838697028745592258709705875168834908314583594102947493349091661801379830005354833653565157987808629144151353148601181998243399
+cipher = 117095409962662175082351896808794751507022710425611245694723238077959955609489552866653772250169627679450018400780468195080203740614508425379908667955368095074633690882446901510961333353751345467769557765781287285368700648236247094896521401160296460446518097963846290488354458277094831166588801569124630935182
+e = 65537
+root = gmpy2.iroot(n,2)[0]
+q = nextprime(root)
+print(n % q)
+p = n // q
+phi = (p - 1) * (q - 1)
+d = inverse(e, phi)
+plaintext = pow(cipher, d, n)
+print(long2str(plaintext))
--- a/xiaosai/factor_n/flag.py
+++ b/xiaosai/factor_n/flag.py
@@ -0,0 +1 @@
+flag = b'flag{You_can_do_this}'
--- a/xiaosai/factor_n/main.py
+++ b/xiaosai/factor_n/main.py
@@ -0,0 +1,16 @@
+from flag import flag
+from Crypto.Util.number import *
+from sympy import nextprime
+
+p = getStrongPrime(512)
+q = nextprime(p)
+n = p * q
+e = 65537
+message = bytes_to_long(flag)
+cipher =  pow(message, e, n)
+print('n = ', n)
+print('cipher = ', cipher)
+'''
+n = 119089874014927074688341672884508166875005676500322264758691717600758737133958853429775347698400246542956663932584349274109281162512900740554148968627493442686180696496512753102593837838697028745592258709705875168834908314583594102947493349091661801379830005354833653565157987808629144151353148601181998243399
+cipher = 117095409962662175082351896808794751507022710425611245694723238077959955609489552866653772250169627679450018400780468195080203740614508425379908667955368095074633690882446901510961333353751345467769557765781287285368700648236247094896521401160296460446518097963846290488354458277094831166588801569124630935182
+'''
				`@@ -0,0 +1 @@`
				`flag = b'flag{W0W_you_cAn_factor_2048bits_prime}'`