1 #include<cstdio>
 2 #include<cmath>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<vector>
 6 #include<stack>
 7 #include<cstring>
 8 #include<queue>
 9 #include<set>
10 #include<string>
11 #include<map>
12 #define inf 9223372036854775807
13 #define INF 9e7+5
14 #define PI acos(-1)
15 using namespace std;
16 typedef long long ll;
17 typedef double db;
18 typedef long long LL;
19 const int maxn = 1e2 + 5;
20 const int mod = 1e9 + 7;
21 const db eps = 1e-9;
22 ll n, l, r, ai[maxn], pi[maxn];
23 
24 int Bitcount(ll x) {
25     return x ? Bitcount(x >> 1) + (x&1LL) : 0;
26 } //求当前状态选择的方程的个数
27 
28 ll q_mul(ll a, ll b, ll mod) {
29     ll ret = 0;
30     while (b) {
31         if (b & 1) ret = (ret + a) % mod;
32         b >>= 1;
33         a = (a + a) % mod;
34     }
35     return ret;
36 } // 快速乘法
37 
38 void Ex_Gcd(ll a, ll b, ll &x, ll &y){
39     ll d;
40     if(b==0){
41         x=1,y=0;
42         return;
43     }
44     Ex_Gcd(b,a%b,y,x);
45     y-=a/b*x;
46 } // 这里直接抄的标程的函数。。
47 
48 ll CRT(ll p[], ll a[], ll cnt) {
49     ll M = 1;
50     ll res = 0;
51     for (int i = 0; i <= cnt; i++) M *= p[i];
52     for (int i = 0; i <= cnt; i++) {
53         ll m = M / p[i], x, y;
54         Ex_Gcd(m, p[i], x, y);
55         res = (res + q_mul(q_mul(x, m, M), a[i], M)) % M;
56     }
57     return (res + M) % M;
58 }
59 
60 ll solve(ll x) {
61     ll p[maxn], m[maxn], ans = 0;
62     p[0] = 7, m[0] = 0;
63     if (!x) return 0;
64     for (ll i = 1; i < (1LL << n); i++) {
65         ll num = Bitcount(i), t = 7, cnt = 0;
66         for (ll j = 0; j < n; j++) {
67             if (i & (1LL << j)) {
68                 p[++cnt] = pi[j];
69                 m[cnt] = ai[j];
70                 t *= pi[j];
71             }
72         }
73         ll res = CRT(p, m, cnt);
74         if (res > x) continue;
75         if (num & 1) ans -= (x - res) / t + 1;  //容斥一下
76         else ans += (x - res) / t + 1;
77     }
78     return ans + x / 7;
79 }
80 
81 int main() {
82     //cin.sync_with_stdio(false);
83     //freopen("tt.txt", "r", stdin);
84     //freopen("isharp.out", "w", stdout);
85     int t, cas = 1;
86 
87     cin >> t;
88     while (t--) {
89         cin >> n >> l >> r;
90         for (int i = 0; i < n; i++)
91             cin >> pi[i] >> ai[i];
92         printf("Case #%d: %I64d\n", cas++, solve(r) - solve(l-1));
93     }
94     return 0;
95 }