1 #include <cstdio>
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <cstring>
 5 #include <queue>
 6 #include <cmath>
 7 using namespace std;
 8 typedef long long LL;
 9 #define N 20
10 
11 LL p[N], a[N];
12 int bit[N];
13 int n;
14 
15 LL mul(LL a, LL b, LL m)
16 {
17     //快速乘法
18     LL ans = 0;
19     while(b) {
20         if(b & 1) ans = (ans + a) % m;
21         a <<= 1;
22         a %= m;
23         b >>= 1;
24     }
25     return ans;
26 }
27 
28 LL exgcd(LL a, LL b, LL &x, LL &y)
29 {
30     if(b == 0) {
31         x = 1;
32         y = 0;
33         return a;
34     }
35     LL r = exgcd(b, a%b, x, y);
36     int t = x;
37     x = y;
38     y = t - a / b * y;
39     return r;
40 }
41 
42 LL CRT(LL x, LL y)
43 {
44     //中国剩余定理: 找同时满足多个同余式的解
45     LL M = 1, ans = 0;
46     for(int i = 0; i <= n; i++) {
47         if(bit[i]) M *= p[i];
48     }
49     for(int i = 0; i <= n; i++) {
50         if(bit[i]) {
51             LL x, y, Mi;
52             Mi = M / p[i];
53             exgcd(Mi, p[i], x, y);
54             x = (x % p[i] + p[i]) % p[i];
55             ans = (ans + mul(Mi * a[i] % M, x, M) % M + M) % M;
56             //ans找出来的是在 M 以内的特解即最小正整数解
57         }
58     }
59     //每过 M 可以有一个解
60     LL res = (y - ans + M) / M - (x - 1 - ans + M) / M;
61     return res;
62 }
63 
64 void solve(LL x, LL y)
65 {
66     bit[n] = 1;
67     LL ans = 0;
68     int all = 1 << n;
69     for(int i = 0; i < all; i++) {
70         int tmp = i, k = 0;
71         for(int j = 0; j < n; j++) {
72             bit[j] = tmp & 1;
73             tmp >>= 1;
74             k += bit[j];
75         }
76         k = k & 1 ? -1 : 1;
77         //k是计算包含多少个同余式
78         //容斥原理: 奇数减,偶数加,具体可以看《组合数学》P108
79         //计算出不具有性质（满足任意一个同余式）的数的数量
80         ans += CRT(x, y) * k;
81     }
82     printf("%I64d\n", ans);
83 }
84 
85 int main()
86 {
87     int t;
88     scanf("%d", &t);
89     for(int cas = 1; cas <= t; cas++) {
90         LL x, y;
91         scanf("%d%I64d%I64d", &n, &x, &y);
92         for(int i = 0; i < n; i++)
93             scanf("%I64d%I64d", &p[i], &a[i]);
94         p[n] = 7, a[n] = 0;
95         printf("Case #%d: ", cas);
96         solve(x, y);
97     }
98     return 0;
99 }