Given two positive integers G and L, could you tell me how many solutions of (x, y, z) there are, satisfying that gcd(x, y, z) = G and lcm(x, y, z) = L?
Note, gcd(x, y, z) means the greatest common divisor of x, y and z, while lcm(x, y, z) means the least common multiple of x, y and z.
Note 2, (1, 2, 3) and (1, 3, 2) are two different solutions.

Input

First line comes an integer T (T <= 12), telling the number of test cases.
The next T lines, each contains two positive 32-bit signed integers, G and L.
It’s guaranteed that each answer will fit in a 32-bit signed integer.

Output

For each test case, print one line with the number of solutions satisfying the conditions above.

Sample Input

2 
6 72 
7 33

Sample Output

72 
0
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=2e5;
bool p[maxn];
int prime[maxn],cnt=0;
typedef long long ll;
void isp(){
	p[0]=p[1]=true;
	for(int i=2;i<maxn;i++){
		if(!p[i]) prime[cnt++]=i;
		for(int j=0;j<cnt&&(ll)i*prime[j]<maxn;j++){
			p[i*prime[j]]=true;
			if(i%prime[j]==0) break;
		}
	}
}
int main(){	
	isp();
	int T;
	scanf("%d",&T);
	while(T--){
		ll G,L;
		scanf("%lld%lld",&G,&L);
		if(L%G) printf("0\n");
		else{
			ll ans=1;
			ll k=L/G;
			for(int i=0;k!=1&&i<cnt;i++){
				int t=0;
				while(k%prime[i]==0){
				    k/=prime[i];
					t++;
					
				}
				if(t>0)
				ans*=6*t;
			}
			if(k!=1) ans*=6*1;
			printf("%lld\n",ans);
		}
	}
	return 0;
}