A

If you think codes, eat codes then sometimes you may get stressed. In your dreams you may see huge codes, as I have seen once. Here is the code I saw in my dream.

#include <stdio.h>

int cases, caseno;
int n, K, MOD;
int A[1001];

int main() {
    scanf("%d", &cases);
    while( cases-- ) {
        scanf("%d %d %d", &n, &K, &MOD);

        int i, i1, i2, i3, ... , iK;

        for( i = 0; i < n; i++ ) scanf("%d", &A[i]);

        int res = 0;
        for( i1 = 0; i1 < n; i1++ ) {
            for( i2 = 0; i2 < n; i2++ ) {
                for( i3 = 0; i3 < n; i3++ ) {
                    ...
                    for( iK = 0; iK < n; iK++ ) {
                        res = ( res + A[i1] + A[i2] + ... + A[iK] ) % MOD;
                    }
                    ...
                }
            }
        }
        printf("Case %d: %d\n", ++caseno, res);
    }
    return 0;
}

Actually the code was about: 'You are given three integers n,K, MOD and n integers: A₀, A₁, A₂ ... A_n-1, you have to writeK nested loops and calculate the summation of all A_i wherei is the value of any nested loop variable.'

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with three integers: n (1 ≤ n ≤ 1000), K (1 ≤ K < 2³¹), MOD (1 ≤ MOD ≤ 35000). The next line containsn non-negative integers denoting A₀, A₁, A₂ ... A_n-1. Each of these integers will be fit into a 32 bit signed integer.

Output

For each case, print the case number and result of the code.

Sample Input

2

3 1 35000

1 2 3

2 3 35000

1 2

Sample Output

Case 1: 6

Case 2: 36

#include <bits/stdc++.h>using namespace std;typedef long long ll;ll n,m,sum,mod;ll quick_pow(ll a,ll n){    ll result =1;    while (n)    {         if (n&1)         result=a*result%mod;         a=a*a%mod;         n>>=1;    }    return result ;}/****//推导过程1.在某一重循环中取得每个数字的概率为k/n2.一共循环n^k,3.n^k(k/n)化简:n^(k-1)*k*****/int main (){      int t,cnt ;      //ll n,k,sum,mod;      cin>>t;      int x;      cnt =1;      ll ans ;      while (t--)      {         sum=0;        cin>>n>>m>>mod;        for (int i=0;i<n;i++)        {             scanf("%d",&x);             sum+=x;        }        if(sum>=mod)        sum%=mod;        ans=quick_pow(n,m-1)*m;        ans=(ans*sum)%mod;        printf("Case %d: %lld\n",cnt++,ans);      }     return 0;}