G - GCD and LCM electrically Hang

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.

InputFirst 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.OutputFor 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 
yesterday like a long time did not figured out this morning ,, sudden inspiration came. 
Solution: First, we must understand that the greatest common divisor and least common multiple of the relationship, such as a * b = gcd (a, b) * lcm (a, b) if the two sides at the same time divided by the square of gcd lcm% gcd == 0 so !, if lcm% gcd = 0, then there should not have relations 
second: we let gcd and lcm gcd available at the same time divided by the new gcd and lcm gcd = 1 lcm = lcm / gcd they are x / gcd y / gcd z / gcd gcd and the lcm lcm we can be decomposed / gcd can be a 
third: max lcm ^ = P1 (A1, A2, A3) * P2 ^ max (B1, B2, B3) .... 
　　　   X0 = P1 a1 .... ^ 
 　　　　Y0 = p1 ^ b1 ... 
　　　　Z0 = p1 ^ c1 ... 
If you say a1 b1 c1 are not 0, then they will not be the greatest common divisor 1, so they have at least three a is 0, a maximum of two to 0 (0 to 3 does not occur, p1 which must be prime factors of a number only) because it is sequential 
(0, a1, C1) add up to less a1 the value of C1 0 - a1 we consider equal to 0 and the situation there in order a1-1 ,, the change to a total of 6 (a1-1) species, as well as (0,0, a1) and (0 , a1, a1) we did not consider a total of 3 + 3 kinds Therefore a total of 6 (a1-1) +6 kinds

#include<iostream>
#include<cstdio>
using namespace std;
const int N=1e6+7;
bool p[N]={1,1,0};
int prime[N];
int k=0;
void pre(){
    k=0;
    for(int i=2;i<N;i++){
        if(p[i]==0){
            prime[k++]=i;
            for(intI = I + J; J <= N; = J + I) { 
                P [I] = . 1 ; 
            } 
        } 
    } 
} 

int main () { 
    pre (); 
    int T; 
    CIN >> T;
     the while (T-- ) {
         int n-, m; 
        Scanf ( " % D% D " , & n-, & m);
         IF (!% n-m = 0 ) { // common divisor should be the most common multiple of the coefficients a * b = gcd example * both sides while divided by the square of gcd lcm, gcd = 0% SO lcm 
            the puts ( " 0 " );
             Continue  ;
        }
        int x=m/n;
        int sum=1;
        for(int i=0;i<k&&prime[i]<x;i++){
            if(x%prime[i]==0){
                int ans=0;
                while(x%prime[i]==0){
                    ans++;
                    x=x/prime[i];
                }
                sum*=6*ans;
            }
        }
        if(x>1) sum*=6;
        cout<<sum<<endl;
        
        
    }
    
    return 0;
}