Problem:
Score : 300 points

Problem Statement

Tak has N cards. On the i-th (1 \leq i \leq N) card is written an integer x_i.
He is selecting one or more cards from these N cards, so that the average of the integers written on the selected cards is exactly A.
In how many ways can he make his selection?

Constraints

1 \leq N \leq 50

1 \leq A \leq 50

1 \leq x_i \leq 50

N,\,A,\,x_i are integers.

Partial Score

200 points will be awarded for passing the test set satisfying 1 \leq N \leq 16.

Input

The input is given from Standard Input in the following format:

N A
x_1 x_2 ... x_N

Output

Print the number of ways to select cards such that the average of the written integers is exactly A.

Sample Input 1

4 8
7 9 8 9

Sample Output 1

5

The following are the 5 ways to select cards such that the average is 8:

Select the 3-rd card.

Select the 1-st and 2-nd cards.

Select the 1-st and 4-th cards.

Select the 1-st, 2-nd and 3-rd cards.

Select the 1-st, 3-rd and 4-th cards.

Sample Input 2

3 8
6 6 9

Sample Output 2

0

Sample Input 3

8 5
3 6 2 8 7 6 5 9

Sample Output 3

19

Sample Input 4

33 3
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

Sample Output 4

8589934591

The answer may not fit into a 32-bit integer.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include<iostream>
#include<cstring>
#include<cstdint>
#include<cstddef>
using namespace std;
using i32 = int_fast32_t;
using i64 = int_fast64_t;
using usize = uint_fast64_t;
#define rep(i, n) for (usize i = 0; i < (usize)(n); i++)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
using P = pair<i64,i64>;
usize dp[55][2501][55];
int main(){
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    i64 n,a;
    cin >> n >> a;
    usize x[n];
    rep(i,n){
        cin >> x[i];
    }
    memset(dp,0,sizeof(dp));
    dp[0][0][0] = 1;
    rep(i,n){
        rep(j,n * a + 1){
            rep(k,n + 1){
               if(dp[i][j][k] == 0)continue;
               dp[i + 1][j][k] += dp[i][j][k];
               dp[i + 1][j + x[i]][k + 1] += dp[i][j][k];
            }
        }
    }
    i64 ans = 0;
    for(i32 i = 1; i <= n; i++){
        ans += dp[n][i * a][i];
    }
    cout << ans << endl;
}