# Fantasy of a Summation

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() { int i, i1, i2, i3, ... , iK; scanf("%d", &cases); while( cases-- ) { scanf("%d %d %d", &n, &K, &MOD); 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[i3] + ... + A[iK] ) % MOD; } ... } } } printf(Case %d: %d\n, ++caseno, res); } return 0; } Actually the code was about: ‘You are given 3 integers n, K, MOD and n integers A0, A1, A2, ..., An−1. You have to write K nested loops and calculate the summation of all Ai where i is the value of any nested loop variable.’ Now you have to find the result according to the code. Input The first line of input contains T denoting the number of cases.

2/2 Each case starts with three integers — n (1 ≤ n ≤ 1000), K (1 ≤ K < 231), MOD (1 ≤ MOD ≤ 35000). The next line will contain n non-negative integers denoting A0, A1, A2, ..., An−1. Each of these integers will be fit into a 32 bit signed integer. Output For each case print the case number and the result. Follow the sample output for the exact output format. Sample Input 2 3 1 35000 123 2 3 35000 12 Sample Output Case 1: 6 Case 2: 36