Ruins is driving a car to participating in a programming contest. As on a very tight schedule, he will drive the car without any slow down, so the speed of the car is non-decrease real number.

Of course, his speeding caught the attention of the traffic police. Police record NN positions of Ruins without time mark, the only thing they know is every position is recorded at an integer time point and Ruins started at 00.

Now they want to know the minimum time that Ruins used to pass the last position.
Input
First line contains an integer TT, which indicates the number of test cases.

Every test case begins with an integers NN, which is the number of the recorded positions.

The second line contains NN numbers a1a1, a2a2, ⋯⋯, aNaN, indicating the recorded positions.

Limits
1≤T≤1001≤T≤100
1≤N≤1051≤N≤105
0

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<math.h>
#include <queue>
#include <functional>
using namespace std;
#define LL long long
const LL maxn = 1e5 + 6;
LL a[maxn];
int main()
{
    LL T, cas = 0;
    scanf
 
("%lld", &T);
    while (T--) {
        LL n, k; scanf("%lld %lld", &n, &k);
        LL sum = 0;
        for (LL i = 1; i <= n; i++) {
            scanf("%lld", &a[i]);
            sum += a[i];
        }
        if (sum%k != 0) {
            printf("Case #%lld: -1\n", ++cas);
            continue;
        }
        LL aim = sum / k;
        LL ans = 0;
        for (LL i = 1; i<n; i++) {
            if (a[i]>aim) {
                LL num = a[i] / aim;
                ans += num;
                if (a[i] % aim == 0) ans--;
                else
                    ans++;
                a[i] %= aim;
                a[i + 1] += a[i];
            }
            else if (a[i]<aim) {
                ans++;
                a[i + 1] += a[i];
            }
        }
        ans += a[n] / aim - 1;
        printf("Case #%lld: %lld\n", ++cas, ans);


    }


    return 0;
}