Description

Fermat's theorem states that for any prime number p and for any integer a > 1, ap = 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 ≤ 1000000000 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
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 typedef long long ll;
 7 ll quick_pow(ll a,ll b,ll mod) {
 8     ll ans=1;
 9     while(b) {
10         if(b&1)
11             ans=(ans*a)%mod;
12         a=(a*a)%mod;
13         b>>=1;
14     }
15     return ans;
16 }
17 bool isprime(int x) {
18     for(int i=2;i*i<=x;i++) {
19         if(x%i==0) return false;
20     }
21     return true;
22 }
23 int main() {
24 //  freopen("input.txt","r",stdin);
25     int a,p;
26     while(scanf("%d%d",&p,&a)!=EOF) {
27         if(a==0&&p==0) break;
28         if(isprime(p)) {
29             printf("no\n");
30             continue;
31 
32         }
33         ll z=quick_pow(a,p,p);
34         if(z==a) printf("yes\n");
35         else printf("no\n");
36     }
37     return 0;
38 }
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