Problem Description

Fermat's theorem states that for any prime number p and for any integer a > 1, a^p == a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a pseudoprimes, have this property for some a. (And some, known as Carmichael Numbers, are base-a pseudoprimes for all a.)
Given 2 < p ≤ 1,000,000,000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.

Input

Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a.

Output

For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".

Sample Input

3 2
10 3
341 2
341 3
1105 2
1105 3
0 0

Sample Output

no
no
yes
no
yes
yes

题意：输入两个数p,a；如果a的p次方对p取余等于a，并且p不是素数，则输出“yes”,否则输出“no”.

这里用到快速幂求余技巧

#include <iostream>

#include <stdio.h>

using namespace std;

bool isprime(long long n){

    for (long long i = ; i*i <= n; i++){

        if (n%i == )

            return false;

    }

    return true;

}

long long qmod(long long a, long long r, long long m){

    long long res = ;

    while (r){

        if (r & )

            res = res*a%m;

        a = a*a%m;

        r >>= ;

    }

    return res;

}

int main(){

    long long p, a;

    while (scanf("%I64d%I64d", &p, &a) && p&&a){

        if (!isprime(p) && qmod(a, p, p) == a)

            printf("yes\n");

        else

            printf("no\n");

    }

    return ;

}