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 p^th 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 ≤ 1000000000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.

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

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

#include<iostream>

#include<cstdio>

#define LL long long

#define N 100000

using namespace std;

int prime[N];

int pn=0;

bool vis[N];

LL pow(LL a,LL n,LL mod)

{

    LL base=a,ret=1;

    while(n)

    {

        if(n&1) ret=(ret*base)%mod;

        base=(base*base)%mod;

        n>>=1;

    }

    return ret%mod;

}

bool judge(int n)

{

    for(int i=0;prime[i]*prime[i]<=n;i++)

    {

        if(n%prime[i]==0)

            return 1;

    }

    return 0;

}

int main()

{

    for (int i = 2; i < N; i++) {

        if (vis[i]) continue;

        prime[pn++] = i;

        for (int j = i; j < N; j += i)

            vis[j] = 1;

    }

    int a,p;

    while(~scanf("%d%d",&p,&a),a&&p)

    {

        if(!judge(p)){

            puts("no");

            continue;

        }

        if(pow(a,p,p)%p==a)

            puts("yes");

        else

            puts("no");

    }

}