1 //合数分解
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 #define ll long long
 7 const int maxn = 1000005;
 8 int prime[maxn];
 9 bool pri[maxn];
10 int cnt;
11 void init()
12 {
13     cnt = 0;
14     pri[0] = pri[1] = 1;
15     //prime[0] = 2;
16     for(int i = 2; i < maxn; i++){
17         if(!pri[i]){
18 
19             prime[cnt++] = i;
20             for(int j = i+i; j < maxn; j+=i)
21             {
22                 pri[j]=1;
23             }
24         }
25     }
26     return;
27 }
28 ll m1[10005],m2[10005];
29 ll c1[10005],c2[10005];
30 int main()
31 {
32     init();
33     int T;
34     ll G,L;
35     scanf("%d",&T);
36     while(T--)
37     {
38         memset(m1,0,sizeof(m1));
39         memset(c1,0,sizeof(c1));
40         memset(m2,0,sizeof(m2));
41         memset(c2,0,sizeof(c2));
42         scanf("%lld%lld",&G,&L);
43         int tm = 0;
44         bool in = 0;
45         bool fl = 1;
46         //printf("%d\n",prime[1]);
47         if(L%G) fl = 0;
48         for(int i = 0; i< cnt; i++){
49             while(prime[i]<=L&&L%prime[i]==0){
50                 L = L/prime[i];
51                 m2[tm] = prime[i];
52                 c2[tm]++;
53                 in = 1;
54             }
55             if(in) tm++;
56             in = 0;
57         }
58         if(L!=1){
59             m2[tm] = L;
60             c2[tm] = 1;
61             tm++;
62         }
63         for(int i = 0; i < tm; i++){
64             while(m2[i]<=G&&G%m2[i]==0){
65                 G = G/m2[i];
66                 m1[i] = m2[i];
67                 c1[i]++;
68             }
69             in = 0;
70         }
71         if(G!=1){
72            fl = 0;
73         }
74         /*puts("haha");
75         for(int i = 0; i < tm; i++){
76             printf("m1[%d]=%d; m2[%d]=%d;\n",i,m1[i],i,m2[i]);
77             printf("c1[%d]=%d; c2[%d]=%d;\n",i,c1[i],i,c2[i]);
78         }
79         */
80         if(fl==0) {puts("0"); continue;}
81         ll ans = 1;
82         for(int i = 0; i < tm; i++){
83             //需要特判当c1[i] =c2[i]的情况
84             ll E;
85             if(c1[i]==c2[i]) E = 0;
86             else E = c2[i]-c1[i]-1;
87             /*
88                 也可以用排列组合的想法，每次找到两个数包含首尾的两个值，然后中间的一个在从所有可能中选取一个。
89                 这样最后的公式就是A(3,2)*(c2[i]-c1[i]-1);
90                 所以
91                 if(c2[i]>c1[i]) ans = ans*6*(c2[i]-c1[i]-1);
92                     */
93             ans = ans*((c2[i]-c1[i]+1)*(c2[i]-c1[i]+1)*(c2[i]-c1[i]+1)-2*(c2[i]-c1[i])*(c2[i]-c1[i])*(c2[i]-c1[i])+E*E*E);
94         }
95         printf("%lld\n",ans);
96     }
97     return 0;
98 }