如何计算D代表从P,Q和E RSA加密 [英] How to calculate D for RSA encryption from P,Q and E
问题描述
我想找到 D
使用 P
,问:
和电子
(压差
,的Dq
和(P
1
模Q)
可太)。
根据这个答案和的这个答案并更新这个问题使用下面的方法,我应该得到 D
。
要测试这个我生成的密钥对,并试图从现有的计算组件和比较的结果与原件。所有的成绩都不错,除了 D
。有什么毛病我的计算,我从上面的答案复制。
这将是巨大的,如果有人能告诉我什么,我做错了。
测试code
使用系统;
使用System.Numerics;
使用System.Security.Cryptography;
使用System.Text;
类节目{
静态RSAParameters键=新RSAParameters(){
P =新的字节[] {
写0xDE,0xA6,0x35,0x0B中,0x0A的,0xA5的,0xD7,0XA0,0x5c的,0x49,0xEA,0xD1,0x3F的,0xA6,0xF5,0×12,
的0x19,0×06,0x25,0x8A,0xD9,0xA7,0×07,0xE7,0X0D,0x8A,0x7C,0xB1,0xD4,0×81,0x64,是0xFD,
0×04,0xEC,0X47,0x33,0x42后,0x0B中,输入0x22,0xF2,地址0x60,为0xBB,0x75,0X62,0x53,0x3E的,0x1A的,0x97,
0x9D,0xEF,0x25,0xA7,为0xE5,0X24,0x3A的为0x30,0x36数据,0xA5的,0xF9,0x8A,0xF5,0xFF时,0x1D,0x1B
},
Q =新的字节[] {
0xBE,0xB9,地址0x60,0×12,0×05,0xB1,0x61,0xD9,输入0x22,为0xD8,的0x84,0x6E,0x9A执行,0x7B,0xD1,0x9B,
0x17已,0xA5的,0xDD,0X02,0x5E,0x9D,为0xD8,0X24,0×06,0x1B,0xF3,为0xD8,值为0x2F,x79的,0xFE时,0x78,
0x74,0x3D之间,0xC4,0xE6,0x17已,0xD2,0xB7,0x68,0x78,0x6F,0x53,取0xE0,0x38,0x00时,0x86可以,0xFB的才能,
为0x20,0x2A,0x1B,0xBD,0x91的信息,0x76,0x3E的,0x33,0x85未,0x9A执行,0X31,0xE6,均为0x88,地址0x60,0x91的信息,0x81表示
},
DP =新的字节[] {
0xAC,0x28,0x92,0x6D,0×46,的0x3F,0x74,0x1A的,0XA0,为0x21,0xDB,为0xBB,为0x0E,0xDF,0xD7,0X31,
0xB6,0x3D之间,0xC5,0x7B,0xB6,0xCE,0x6B,0xD2,0xE1,0xEA,0x8A,0x7E的,和0xAA,0xD5,0x9E,0xB3,
0xF2,0×41,0x8C,0xD0,0x7A,0xA9,0xC7,的0xCC,0xE8,0xB5执行,0x2A,0x8F,将0xEB,0xD3,0xE2,0x96,
0×07,0xDD,0xEA,0x1D,0×07,0x96,5AH即可,0x93,0xFB的才能,0x3D之间,0x9D,0x56,为0x30,写0xDE,0xA1,0xAF执行
},
DQ =新的字节[] {
0xA6,为0x9c,0x44,0x1B,0x9A执行,0x53,0x89上,0xD9,0xE8,0xC1,0xE2,0x76,0xC8,87H的,0x6F,为0xE5,
为0x1F,0x74,的0x6A,0xAC,0x5E,0×41,0x5F的,0x86可以,0XA0,为0xBB,为0x9c,x79的,0xF7,87H的,87H的,0xD0,
0x6C,0x23,0x65,0xB5执行,0x67,0x8C,0x51,0X62,0x77,0x0B中,0X31,0xE7,0x86可以,0xA4,0x97,0×46,
0x1B,0xA4,0X0D,写入0x55,0xBE,0x13,取0xE0,0x64,0x9B,0xCA,0xC6,0xDA,0xCF,0xBA,0X24,0x81表示
},
InverseQ =新的字节[] {
0×02,的0x42,0x90处,0xAE,为0xFF,0xFE的,0xB6,0xCB,0x53,0xFF时,0x96,0x17已,0xC6,0xE4,0x3F的,0xE6,
0xC7,0xBC,0xB2,将0xEB,0x53,0xA9,0X47,0xEE,为0x10,0x36数据,0x98在全局,0xEF,0xA8,0x3E的,为0x9c,0xF7,
0xF9,0xCF,0X24,为0xE5,0xD7,0x9A执行,0xAF执行,×09,0xCF,0x28,和0xAA,0x5D,0x2A,0xB7,0x27(0x73)的,
0X47,0x2D,0x54,0x54,0x61,0xC5,0xCE,0x3E的,0xA4,0x91的信息,0xF6,0x9D,0xF4中,0x65,0x08的,0xDD
},
指数=新的字节[] {
0×00,0×01,0×00,0×01,
},
模量=新的字节[] {
0xA5的,取0xE0,位0x95,0x08的,87H的,0×69,0x2B访问,0xB4,0x7F的,0x08的,0xFB的才能,0x4F,0x66,0x85未,0xD9,位0x95,
0x53,为0x0F,0x7C,0x99,位0x95,为0x16,0xF4中,0X0D,写入0xAD,0x9E,0X31,为0xD8,为0x20,0xF4中,均为0x88,0x63,
0xAE,0x51,0x04的为0xC2,0xE9,0x92,0x3C符号,为0x1C,0×90,0xF8时,0xF4中,0x38,的0x6A,0x86可以,是0xFD,0x8F,
写0xDE,0x85未,输入0x22,0xDD,0xE8,0x7E的,0x8D,0xF2,0xC5,0xC9,0x4E,0x71,0x2B访问,0x56,0x25,0x1A的,
0xEA,0x66,0x15,的0x19,0x63,0x70,0x53,x79的,0xDF,0x38,0x49,为0x30,0x74,×45,0xBE,0xA3执行,
0x28,0X0D,为0x0E,0x7A,0x7D,0xB6,0x8B,0xCA,×09,0x56,0×21,0xE7,0x98在全局,0x3E的,0x4B,0x8B,
0xD0,0X31,0x27,为0x8E,0x6F,0x10的,0xA6,0x6C,为0x1C,0x48,0xB5执行,0x5E,0x89上,0x7B,0x74,0x74,
0xB2,0×57,0x72,0x6D,为0x18,将0xEB,0xF3,0xF5,0x53,0xCA,0x8C,0xBE,0xB7,0x29,0xF5,0x9B
},
D =新的字节[] {
0x9F,0x86可以,0xE1,送出0x4d,0x96,0x8C,0xFA回应,0xCF,0×57,0xED,0x17已,0x64,0x41既,0×41,0X31,0×04,
0x7F的,0×21,×41,为0xBF,0xA2,0xB6,0xB4,0x78,0×03,0x25,0x44,0xE2,0x8A,0xAF执行,输入0x22,0x0C的,
0x5B,0xB4,0xE7,0x53,0x5c的,0xB6,0x9A执行,0xC1,为0x0E,0x5B,0x9E,0xE4,0x32,0xEF,0x28,0X24,
0x98在全局,0xE8,0x89上,0xA3执行,0xC8,0xD9,0X0D,0×43,×12,为0x1C,0x8C,0x28,输入0x22,x79的,0x72,0xAC,
0x66,0x7B,0x7D,0xD2,0xF9,0x48,0×06,0XCD,0x9D,0x9A执行,0xE6,0x42后,0x92,0xBA,0x56,0xA6,
0x63,0x07执行,0X1E,0x25,0x4E,0xC8,0x07执行,将0x58,0x5B,均为0x88,地址0x60,0x97,0x92,0xE2,0xD5,0xB9,
0xC6,0x70,为0xBB,0x63,5AH即可,0xC3,0xC3,0xA6,0×46,5AH即可,为0x1C,为0x9c,为0xBF,0x61,0×57,0x9E,
0x9E,0xFA回应,为0xC0,0xC4,0x8A,为0xC2,0xBA,均为0x88,将0x46,0xA9,0x7A,0xF2,0x7D,0x4F,0x6C,0×01
}
};
公共静态的BigInteger FromBigEndian(byte []的P){
Array.Reverse(对);
如果(对[p.Length - 1]≥127){
Array.Resize(参考磷,p.Length + 1);
P [p.Length - 1] = 0;
}
返回新的BigInteger(P);
}
静态无效的主要(字串[] args){
使用(的RSACryptoServiceProvider RSA =新的RSACryptoServiceProvider(){PersistKeyInCsp = FALSE}){
rsa.ImportParameters(键);
Console.Write(测试加密/解密......);
字符串消息=测试一些加密数据;
byte []的缓冲区= Encoding.ASCII.GetBytes(消息);
byte []的EN codeD = rsa.Encrypt(缓冲,真正的);
byte []的德codeD = rsa.Decrypt(EN codeD,真正的);
字符串MESSAGE1 = ASCIIEncoding.ASCII.GetString(德codeD);
如果(消息== MESSAGE1){
Console.WriteLine(好:));
} 其他 {
Console.WriteLine(坏加密:();
Console.ReadKey();
返回;
}
}
//转换关键BigIntegers
BigInteger的P = FromBigEndian(key.P);
BigInteger的Q = FromBigEndian(key.Q);
BigInteger的DP = FromBigEndian(key.DP);
BigInteger的DQ = FromBigEndian(key.DQ);
BigInteger的InverseQ = FromBigEndian(key.InverseQ);
BigInteger的E = FromBigEndian(key.Exponent);
BigInteger值M = FromBigEndian(key.Modulus);
BigInteger的D = FromBigEndian(key.D);
Console.WriteLine(测试号......);
的BigInteger M1 = BigInteger.Multiply(P,Q); // M = P * Q
如果(M1.CompareTo(M)== 0){
Console.WriteLine(M好:));
} 其他 {
Console.WriteLine(坏中号:();
Console.ReadKey();
返回;
}
BigInteger的PMinus1 = BigInteger.Subtract(P,BigInteger.One); // M = P * Q
BigInteger的DP1 = BigInteger.Remainder(D,PMinus1); // M = P * Q
如果(DP1.CompareTo(DP)== 0){
Console.WriteLine(DP好:));
} 其他 {
Console.WriteLine(坏DP :();
Console.ReadKey();
返回;
}
BigInteger的QMinus1 = BigInteger.Subtract(Q,BigInteger.One); // M = P * Q
BigInteger的DQ1 = BigInteger.Remainder(D,QMinus1); // M = P * Q
如果(DQ1.CompareTo(DQ)== 0){
Console.WriteLine(DQ好:));
} 其他 {
Console.WriteLine(坏DQ :();
Console.ReadKey();
返回;
}
BigInteger的披= BigInteger.Multiply(PMinus1,QMinus1);
BigInteger的PhiMinus1 = BigInteger.Subtract(披,BigInteger.One);
BigInteger的D1 = BigInteger.ModPow(E,PhiMinus1,披);
如果(D1.CompareTo(D)== 0){
Console.WriteLine(D好:));
} 其他 {
Console.WriteLine(出错的D :();
Console.ReadKey();
返回;
}
Console.ReadKey();
}
}
测试结果
测试加密/解密...好:)
测试号码...
中号好了:)
DP好:)
DQ好:)
出错的D :(
首先,你需要确认 GCD(E,φ)= 1
,因为ð
只有在该属性保存存在。然后计算电子
模<$ C的模反元素$ C>披 我描述我在C#1 / BigInteger的答案。
您code似乎认为 E ^(φ(N)-1)模φ(N)
是逆,但是这是不正确。我认为正确的公式应该为 E ^(φ(φ(N)) - 1)模φ(N)
,但这是使用不方便,因为你只知道φ(N)
而不是φ(φ(N))
。
我建议使用扩展欧几里德算法被移植维基百科的伪code到C#。
作为一个侧面说明:的有用于 D
,因为你不需要电子邮件往往是多重的等效值* D模φ(N)= 1
只是 * D + Emodλ(N)= 1
其中λ是的卡迈克尔功能看到<一href="http://crypto.stackexchange.com/questions/1789/why-rsa-encryption-key-is-based-on-modulophin-rather-than-modulo-n">"Why RSA加密密钥基于模(披(n))的,而不是在crypto.SE 的模N
I am trying to find D
using P
,Q
and E
(Dp
, Dq
and (p
-1
mod q)
are available too).
According to this answer and this answer and update for this question using following method I should get D
.
To test this I generated Key pair and tried to calculate components from existing ones and compare the result with originals. All the results are good except for D
. there is something wrong with my calculation which I copied from above answers.
it would be great if someone can tell me what I'm doing wrong.
Test Code
using System;
using System.Numerics;
using System.Security.Cryptography;
using System.Text;
class Program {
static RSAParameters key = new RSAParameters() {
P = new byte[]{
0xDE, 0xA6, 0x35, 0x0B, 0x0A, 0xA5, 0xD7, 0xA0, 0x5C, 0x49, 0xEA, 0xD1, 0x3F, 0xA6, 0xF5, 0x12,
0x19, 0x06, 0x25, 0x8A, 0xD9, 0xA7, 0x07, 0xE7, 0x0D, 0x8A, 0x7C, 0xB1, 0xD4, 0x81, 0x64, 0xFD,
0x04, 0xEC, 0x47, 0x33, 0x42, 0x0B, 0x22, 0xF2, 0x60, 0xBB, 0x75, 0x62, 0x53, 0x3E, 0x1A, 0x97,
0x9D, 0xEF, 0x25, 0xA7, 0xE5, 0x24, 0x3A, 0x30, 0x36, 0xA5, 0xF9, 0x8A, 0xF5, 0xFF, 0x1D, 0x1B
},
Q = new byte[]{
0xBE, 0xB9, 0x60, 0x12, 0x05, 0xB1, 0x61, 0xD9, 0x22, 0xD8, 0x84, 0x6E, 0x9A, 0x7B, 0xD1, 0x9B,
0x17, 0xA5, 0xDD, 0x02, 0x5E, 0x9D, 0xD8, 0x24, 0x06, 0x1B, 0xF3, 0xD8, 0x2F, 0x79, 0xFE, 0x78,
0x74, 0x3D, 0xC4, 0xE6, 0x17, 0xD2, 0xB7, 0x68, 0x78, 0x6F, 0x53, 0xE0, 0x38, 0x00, 0x86, 0xFB,
0x20, 0x2A, 0x1B, 0xBD, 0x91, 0x76, 0x3E, 0x33, 0x85, 0x9A, 0x31, 0xE6, 0x88, 0x60, 0x91, 0x81
},
DP = new byte[]{
0xAC, 0x28, 0x92, 0x6D, 0x46, 0x3F, 0x74, 0x1A, 0xA0, 0x21, 0xDB, 0xBB, 0x0E, 0xDF, 0xD7, 0x31,
0xB6, 0x3D, 0xC5, 0x7B, 0xB6, 0xCE, 0x6B, 0xD2, 0xE1, 0xEA, 0x8A, 0x7E, 0xAA, 0xD5, 0x9E, 0xB3,
0xF2, 0x41, 0x8C, 0xD0, 0x7A, 0xA9, 0xC7, 0xCC, 0xE8, 0xB5, 0x2A, 0x8F, 0xEB, 0xD3, 0xE2, 0x96,
0x07, 0xDD, 0xEA, 0x1D, 0x07, 0x96, 0x5A, 0x93, 0xFB, 0x3D, 0x9D, 0x56, 0x30, 0xDE, 0xA1, 0xAF
},
DQ = new byte[]{
0xA6, 0x9C, 0x44, 0x1B, 0x9A, 0x53, 0x89, 0xD9, 0xE8, 0xC1, 0xE2, 0x76, 0xC8, 0x87, 0x6F, 0xE5,
0x1F, 0x74, 0x6A, 0xAC, 0x5E, 0x41, 0x5F, 0x86, 0xA0, 0xBB, 0x9C, 0x79, 0xF7, 0x87, 0x87, 0xD0,
0x6C, 0x23, 0x65, 0xB5, 0x67, 0x8C, 0x51, 0x62, 0x77, 0x0B, 0x31, 0xE7, 0x86, 0xA4, 0x97, 0x46,
0x1B, 0xA4, 0x0D, 0x55, 0xBE, 0x13, 0xE0, 0x64, 0x9B, 0xCA, 0xC6, 0xDA, 0xCF, 0xBA, 0x24, 0x81
},
InverseQ = new byte[]{
0x02, 0x42, 0x90, 0xAE, 0xFF, 0xFE, 0xB6, 0xCB, 0x53, 0xFF, 0x96, 0x17, 0xC6, 0xE4, 0x3F, 0xE6,
0xC7, 0xBC, 0xB2, 0xEB, 0x53, 0xA9, 0x47, 0xEE, 0x10, 0x36, 0x98, 0xEF, 0xA8, 0x3E, 0x9C, 0xF7,
0xF9, 0xCF, 0x24, 0xE5, 0xD7, 0x9A, 0xAF, 0x09, 0xCF, 0x28, 0xAA, 0x5D, 0x2A, 0xB7, 0x27, 0x73,
0x47, 0x2D, 0x54, 0x54, 0x61, 0xC5, 0xCE, 0x3E, 0xA4, 0x91, 0xF6, 0x9D, 0xF4, 0x65, 0x08, 0xDD
},
Exponent = new byte[]{
0x00, 0x01, 0x00, 0x01,
},
Modulus = new byte[]{
0xA5, 0xE0, 0x95, 0x08, 0x87, 0x69, 0x2B, 0xB4, 0x7F, 0x08, 0xFB, 0x4F, 0x66, 0x85, 0xD9, 0x95,
0x53, 0x0F, 0x7C, 0x99, 0x95, 0x16, 0xF4, 0x0D, 0xAD, 0x9E, 0x31, 0xD8, 0x20, 0xF4, 0x88, 0x63,
0xAE, 0x51, 0x04, 0xC2, 0xE9, 0x92, 0x3C, 0x1C, 0x90, 0xF8, 0xF4, 0x38, 0x6A, 0x86, 0xFD, 0x8F,
0xDE, 0x85, 0x22, 0xDD, 0xE8, 0x7E, 0x8D, 0xF2, 0xC5, 0xC9, 0x4E, 0x71, 0x2B, 0x56, 0x25, 0x1A,
0xEA, 0x66, 0x15, 0x19, 0x63, 0x70, 0x53, 0x79, 0xDF, 0x38, 0x49, 0x30, 0x74, 0x45, 0xBE, 0xA3,
0x28, 0x0D, 0x0E, 0x7A, 0x7D, 0xB6, 0x8B, 0xCA, 0x09, 0x56, 0x21, 0xE7, 0x98, 0x3E, 0x4B, 0x8B,
0xD0, 0x31, 0x27, 0x8E, 0x6F, 0x10, 0xA6, 0x6C, 0x1C, 0x48, 0xB5, 0x5E, 0x89, 0x7B, 0x74, 0x74,
0xB2, 0x57, 0x72, 0x6D, 0x18, 0xEB, 0xF3, 0xF5, 0x53, 0xCA, 0x8C, 0xBE, 0xB7, 0x29, 0xF5, 0x9B
},
D = new byte[]{
0x9F, 0x86, 0xE1, 0x4D, 0x96, 0x8C, 0xFA, 0xCF, 0x57, 0xED, 0x17, 0x64, 0x41, 0x41, 0x31, 0x04,
0x7F, 0x21, 0x41, 0xBF, 0xA2, 0xB6, 0xB4, 0x78, 0x03, 0x25, 0x44, 0xE2, 0x8A, 0xAF, 0x22, 0x0C,
0x5B, 0xB4, 0xE7, 0x53, 0x5C, 0xB6, 0x9A, 0xC1, 0x0E, 0x5B, 0x9E, 0xE4, 0x32, 0xEF, 0x28, 0x24,
0x98, 0xE8, 0x89, 0xA3, 0xC8, 0xD9, 0x0D, 0x43, 0x12, 0x1C, 0x8C, 0x28, 0x22, 0x79, 0x72, 0xAC,
0x66, 0x7B, 0x7D, 0xD2, 0xF9, 0x48, 0x06, 0xCD, 0x9D, 0x9A, 0xE6, 0x42, 0x92, 0xBA, 0x56, 0xA6,
0x63, 0x07, 0x1E, 0x25, 0x4E, 0xC8, 0x07, 0x58, 0x5B, 0x88, 0x60, 0x97, 0x92, 0xE2, 0xD5, 0xB9,
0xC6, 0x70, 0xBB, 0x63, 0x5A, 0xC3, 0xC3, 0xA6, 0x46, 0x5A, 0x1C, 0x9C, 0xBF, 0x61, 0x57, 0x9E,
0x9E, 0xFA, 0xC0, 0xC4, 0x8A, 0xC2, 0xBA, 0x88, 0x46, 0xA9, 0x7A, 0xF2, 0x7D, 0x4F, 0x6C, 0x01
}
};
public static BigInteger FromBigEndian(byte[] p) {
Array.Reverse(p);
if (p[p.Length - 1] > 127) {
Array.Resize(ref p, p.Length + 1);
p[p.Length - 1] = 0;
}
return new BigInteger(p);
}
static void Main(string[] args) {
using (RSACryptoServiceProvider rsa = new RSACryptoServiceProvider() { PersistKeyInCsp = false }) {
rsa.ImportParameters(key);
Console.Write("Testing Encrypt/Decrypt ... ");
string message = "Testing Some Data to Encrypt";
byte[] buffer = Encoding.ASCII.GetBytes(message);
byte[] encoded = rsa.Encrypt(buffer, true);
byte[] decoded = rsa.Decrypt(encoded, true);
string message1 = ASCIIEncoding.ASCII.GetString(decoded);
if (message == message1) {
Console.WriteLine("Ok :)");
} else {
Console.WriteLine("Bad Encryption :(");
Console.ReadKey();
return;
}
}
//Convert Key to BigIntegers
BigInteger P = FromBigEndian(key.P);
BigInteger Q = FromBigEndian(key.Q);
BigInteger DP = FromBigEndian(key.DP);
BigInteger DQ = FromBigEndian(key.DQ);
BigInteger InverseQ = FromBigEndian(key.InverseQ);
BigInteger E = FromBigEndian(key.Exponent);
BigInteger M = FromBigEndian(key.Modulus);
BigInteger D = FromBigEndian(key.D);
Console.WriteLine("Testing Numbers ... ");
BigInteger M1 = BigInteger.Multiply(P, Q); // M = P*Q
if (M1.CompareTo(M) == 0) {
Console.WriteLine(" M Ok :)");
} else {
Console.WriteLine(" Bad M:(");
Console.ReadKey();
return;
}
BigInteger PMinus1 = BigInteger.Subtract(P, BigInteger.One); // M = P*Q
BigInteger DP1 = BigInteger.Remainder(D, PMinus1); // M = P*Q
if (DP1.CompareTo(DP) == 0) {
Console.WriteLine(" DP Ok :)");
} else {
Console.WriteLine(" Bad DP :(");
Console.ReadKey();
return;
}
BigInteger QMinus1 = BigInteger.Subtract(Q, BigInteger.One); // M = P*Q
BigInteger DQ1 = BigInteger.Remainder(D, QMinus1); // M = P*Q
if (DQ1.CompareTo(DQ) == 0) {
Console.WriteLine(" DQ Ok :)");
} else {
Console.WriteLine(" Bad DQ :(");
Console.ReadKey();
return;
}
BigInteger Phi = BigInteger.Multiply(PMinus1, QMinus1);
BigInteger PhiMinus1 = BigInteger.Subtract(Phi, BigInteger.One);
BigInteger D1 = BigInteger.ModPow(E, PhiMinus1, Phi);
if (D1.CompareTo(D) == 0) {
Console.WriteLine(" D Ok :)");
} else {
Console.WriteLine(" Bad D :(");
Console.ReadKey();
return;
}
Console.ReadKey();
}
}
Test Result
Testing Encrypt/Decrypt ... Ok :)
Testing Numbers ...
M Ok :)
DP Ok :)
DQ Ok :)
Bad D :(
First you need to verify that GCD(e, φ) = 1
because d
only exists if that property holds. Then calculate the modular multiplicative inverse of e
modulo phi
which I describe in my answer to "1/BigInteger in C#".
Your code seems to assume that e^(φ(n)-1) mod φ(n)
is that inverse, but that's incorrect. I think the correct formula would be e^(φ(φ(n))-1) mod φ(n)
, but that's inconvenient to use since you only know φ(n)
but not φ(φ(n))
.
I recommend using the Extended Euclidean algorithm by porting the wikipedia pseudocode to C#.
As a side-note: There are often multiple equivalent values for d
since you don't need e*d mod φ(n)=1
but just e*d mod λ(n)=1
where λ is the Carmichael function see "Why RSA encryption key is based on modulo(phi(n)) rather than modulo n" on crypto.SE
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