#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
int cnt, prime[maxn], mu[maxn], vis[maxn];
inline void input(int &x) {
    char c;
    x = 0;
    while ((c = getchar()) < '0' || c > '9');
    while ('0' <= c && c <= '9') x = x * 10 + (c-'0'), c = getchar();
}
void init() {
    int N = maxn;
    memset(prime, 0, sizeof(prime));
    memset(mu, 0, sizeof(mu));
    memset(vis, 0, sizeof(vis));
    mu[1] = 1;
    cnt = 0;
    for (int i = 2; i < N; i++) {
        if (!vis[i]) {
            prime[cnt++] = i;
            mu[i] = -1;
        }
        for (int j = 0; j < cnt && i * prime[j] < N; j++) {
            vis[i*prime[j]] = 1;
            if (i % prime[j]) mu[i*prime[j]] = -mu[i];
            else {
                mu[i*prime[j]] = 0;
                break;
            }
        }
    }
}
int main() {
    init();
    int T, a, b, c, d, k;
    input(T);
    for (int kase = 1; kase <= T; kase++) {
        input(a); input(b); input(c); input(d); input(k);
        if (k == 0) { printf("Case %d: 0\n", kase); continue; }
        ll ans1 = 0, ans2 = 0;
        int n = b/k, m = d/k, t= min(n, m);
        for (int i = 1; i <= t; i++) ans1 += (ll)mu[i]*(n/i)*(m/i);
        for (int i = 1; i <= t; i++) ans2 += (ll)mu[i]*(t/i)*(t/i);
        printf("Case %d: %lld\n", kase, ans1 - ans2/2);
    }
    return 0;
}