1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 typedef long long ll;
 5 const ll maxn = 1e6 +7;
 6 bool isp[maxn];
 7 int pis[maxn], tot;
 8 
 9 void getp(ll n) {//得到n以內的素數及個數
10     memset(isp, true, sizeof(isp));
11     for(ll i = 2; i <= n; i++) {
12         if
 
(isp[i]) {
13             tot++;
14             pis[tot] = i;
15         }
16         for(int j = 1; (j <= tot) && (i * pis[j] <= n); j++) {
17             isp[i * pis[j]] = false;
18             if(i % pis[j] == 0) break;
19         }
20     }
21 }
22 ll fac(ll n) {//計算n的正因子個數之和
23     ll now = n;
 
24     ll ans = 1;
25     for(ll i = 1; i <= tot; i++) {
26         if(pis[i] > now)//重要剪枝，每次不必全部試除一遍才結束
27             break;
28         if(now % pis[i] == 0) {
29             int cnt = 0;
30             while(now % pis[i] == 0) {
31                 cnt++;
32                 now /= pis[i];
33             }
 
34             ans *= (cnt + 1);
35         }
36     }
37     if(now != 1)
38         ans *= 1 + 1;
39     return ans;
40 }
41 ll solve(ll S, ll b) {
42     if(b * b >= S)
43         return 0;
44 
45     ll ans = fac(S);//得到S的正因子個數之和
46     ans /= 2;
47     for(ll i = 1; i < b; i++) {
48         if(S % i == 0)
49             ans--;
50     }
51     return ans;
52 }
53 int main()
54 {
55     ll tot = 0;
56     getp(maxn - 7);//得到maxn - 7以內的素數及個數
57     int T, k = 1;
58     scanf("%d", &T);
59     while(T--) {
60         ll a, b;
61         scanf("%lld%lld", &a, &b);
62         printf("Case %d: %lld\n", k++, solve(a, b));
63     }
64     return 0;
65 }