#include <stdio.h>

#include <string.h>

#include <iostream>

using namespace std;

#define LL long long

#define FOR(i, n) for (int i=0; i<n; ++i)

LL l, r;

int extend_gcd(LL a, LL b, LL &x, LL &y) {

    if (b == 0) {

        x = 1;

        y = 0;

        return a;

    } else {

        int r = extend_gcd(b, a%b, y, x);

        y -= x*(a/b);

        return r;

    }

}

LL qMul(LL a, LL b, LL mod) {

    a %= mod;

    LL ret = 0;

    while(b) {

        if (b & 1) ret = (ret + a) % mod;

        b >>= 1;

        a = (a + a) % mod;

    }

    return ret;

}

LL CRT(LL *a, LL *m, int n) {

    LL M = 1;

    for (int i=0; i<n; ++i) M *= m[i];

    LL x = 0;

    LL d, y;

    for (int i=0; i<n; ++i) {

        //LL d, y;

        LL tm = M/m[i];

        extend_gcd(m[i], tm, d, y);

        x = (x + qMul( qMul(d, tm, M), a[i], M)) % M;  // 参数传递有顺序

    }

    x = (x + M) % M;

    return (r+M-x)/M - (l-1+M-x)/M; // 直接返回l r区间有多少个解。

}

LL a[20], m[20];

LL chu[20], rema[20];

int main() {

    freopen("1.in.cpp", "r", stdin);

    int t;

    scanf("%d", &t);

    int cas = 0;

    while(t--) {

        int n;

        scanf("%d%lld%lld", &n, &l, &r);

        FOR(i, n) {

            scanf("%lld%lld", &chu[i], &rema[i]);

        }

        LL tot = (1<<n);

        LL ans = r/7 - (l-1)/7;

        for (int i=1; i<tot; ++i) {

            //cout << i << "++++\n";

            int cnt = 0;

            FOR (j, n) {

                if (i&(1<<j)) {

                    m[cnt] = chu[j];

                    a[cnt] = rema[j];

                    cnt++;

                }

            }

            m[cnt] = 7, a[cnt] = 0;

            cnt++;

            LL tans = CRT(a, m, cnt);

            cnt = (cnt&1) ? 1 : -1;

            ans += tans * cnt;

        }

        //cout << "++++\n";

        printf("Case #%d: %I64d\n", ++cas, ans);

    }

    return 0;

}