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 ", ++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 write K nested loops and calculate the summation of all A_i where i 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 contains n 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

Output for Sample Input

2

3 1 35000

1 2 3

2 3 35000

1 2

Case 1: 6

Case 2: 36

分析：由题得，程序执行了n^k次加法，每次取k个数，并且每个数出现的次数是相等的为n^k * k/n次。

sum = (a0+a1+...an)%mod;

ans = (sum+k*n^k-1)%mod;


结合律	((a+b) mod p + c)mod p = (a + (b+c) mod p) mod p ((ab) mod p c)mod p = (a * (b*c) mod p) mod p
交换律	(a + b) mod p = (b+a) mod p (a × b) mod p = (b × a) mod p
分配律	((a +b)mod p × c) mod p = ((a × c) mod p + (b × c) mod p) mod p (a×b) mod c=(a mod c * b mod c) mod c (a+b) mod c=(a mod c+ b mod c) mod c (a-b) mod c=(a mod c- b mod c) mod c

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;
typedef long long ll;

#define N 11000
int mod;

ll qpow(int a, int b)
{
    if(b == 0)
        return 1;

    ll tmp = qpow(a, b>>1);

    tmp = tmp * tmp % mod;

    if(b & 1)
       tmp =(ll) (tmp * (a % mod) )% mod;

    return tmp;

}

int main()
{
    int T, cas;
    int n, k;
    int num[N];

    scanf("%d", &T);

    cas = 0;

    while(T--)
    {
        cas++;
        scanf("%d%d%d", &n, &k, &mod);

        ll sum = 0;
        for(int i = 0; i < n; i++)
        {
             scanf("%d", &num[i]);

             sum = (sum + num[i] % mod) % mod;
        }

        ll ans = qpow(n, k-1);
        ll cnt = (ans % mod * k % mod * sum % mod) % mod;

        printf("Case %d: %lld ", cas, cnt);

    }
    return 0;
}

1213

Input

Output

Sample Input

Output for Sample Input