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#define _CRT_SECURE_NO_WARNINGS 1//jerry99的数列
  2 #include<bits/stdc++.h>
  3 int prime[40000] = { 0 };
  4 int pri[6000] = { 0 };
  5 int nums[3000] = { 0 };
  6 int mem[3000] = { 0 };
  7 int jiecheng[40000*1000] = { 0 };
  8 void pre() {//标记质数1~50000
  9     prime[0] = prime[1] = 1;
 10     for (int i = 2; i < 40000; i++) {
 11         for (int j = 2 * i; j < 40000; j += i) {
 12             prime[j] = 1;
 13         }
 14     }
 15     int k = 0;
 16     for (int i = 0; i < 40000; i++) {
 17         if (!prime[i])
 18             pri[k++] = i;//存放在num数组里备用
 19     }
 20 }
 21 
 22 
 23 int div2(int n) {
 24     memset(mem, 0, 6000);
 25     int k = 0, d, num = 0, origin = n;
 26     for (int i = 0; n != 1; i++) {
 27         d = pri[i];
 28         while (n % d == 0) {
 29             mem[k]++;
 30             n /= d;
 31         }
 32         k++;
 33         if (k >= 5133&&n>50000) return 1;
 34     }
 35     return 0;
 36 }
 37 int main() {
 38     pre();//筛出质数
 39 //提前分解好1~20000阶乘质因数
 40 
 41 
 42     int t, n, m;
 43     for (n = 1; n < 40000; n++) {
 44         int k = 0, d, num = 0, origin = n;
 45         for (int i = 0; pri[i]<=n; i++) {
 46             d = pri[i];
 47             while (origin != 1&&origin%d==0) {
 48                 nums[k]++;
 49                 origin /= d;
 50             }
 51             k++;
 52         }
 53         for (int p = 0; p < k; p++) {
 54             *(jiecheng+1000*n+p) += (*(jiecheng+1000*(n-1)+p)+ nums[p]);//递推地计算质因数分解结果
 55             //printf("%d ", jiecheng[n][p]);//数组一维化
 56         }//printf("\n");
 57          memset(nums, 0, 3000*4);
 58     }
 59     for (int i = 1; i < 10; i++) {
 60         for (int j = 0; j < 10; j++) {
 61             printf("%d ", jiecheng[i][j]);
 62         }printf("\n");
 63     }
 64     scanf("%d", &t);
 65     int judge;
 66     while (t > 0) {
 67         memset(mem, 0, 3000*4);
 68         scanf("%d%d", &n, &m);//n分母,m分子
 69         judge = div2(n);//把分母质因数分解
 70 
 71         if (judge&&n>40000) {//如果n分解后的质因数是个超级大(>50000)的质数
 72             if (m < n) printf("0\n");
 73             else {
 74                 printf("%d\n", m - n + 1);
 75             }
 76         }
 77         else {//小于50000的质数或者不是质数的10^9以内的
 78             int cnt, j = 0;
 79             for (cnt = 1; cnt < n;) {
 80                 //printf("%d ", *(jiecheng + 1000 * cnt + j));
 81                 while (mem[j] > * (jiecheng + 1000 * cnt + j)) {
 82                     cnt++;//每一位幂次进行比较
 83                 }
 84 
 85                 //printf("%d ", *(jiecheng + 1000 * cnt + j));
 86                 if (mem[j] <= *(jiecheng + 1000 * cnt + j)) {
 87                     j++;//比较完一个每次进入下一个幂次
 88                 }
 89                 if (!mem[j] && j == n) {
 90                     break;
 91                 }
 92 
 93 
 94             }
 95             /*for (int i = 1; i < n; i++) {
 96                 for (int j =0 ; j < 10; j++) {
 97                     printf("%d ", *(jiecheng +1000 * i + j));
 98                 }
 99             }*/
100             //printf("比%d大的最小阶乘是 %d !\n", n, cnt);
101             if (cnt > m + 1) printf("0\n");
102             else printf("%d\n", m - cnt + 1);
103             t--;
104         }
105     }
106     return 0;
107 }
108