/**

题目：Fantasy of a Summation

链接：https://vjudge.net/contest/154246#problem/L

题意：n个数，k层重复遍历相加。求它的和%mod的值；

思路：很容易想到n个数出现在要加的和中的次数相同。

又所有数的出现次数为n^k * k:

所以每个数出现的次数为n^k * k / n;

*/

#include<iostream>

#include<cstring>

#include<cstdio>

#include<algorithm>

#include<cmath>

using namespace std;

typedef long long ll;

const int maxn = ;

int mod;

int a[maxn];

int n, k;

ll solve(ll a,ll b)

{

    ll p = ;

    while(b>){

        if(b&) p = p*a%mod;

        a = a*a%mod;

        b >>= ;

    }

    return p;

}

int main()

{

    int T, cas=;

    cin>>T;

    while(T--)

    {

        scanf("%d%d%d",&n,&k,&mod);

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

        ll sum = ;

        for(int i = ; i < n; i++) sum += a[i];

        sum%=mod;

        printf("Case %d: %lld\n",cas++,solve(n,k-)*k%mod*sum%mod);

    }

    return ;

}