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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

// finish date: 2018/01/27
#include <iostream>
#include <cmath>
#include <vector>
#include <bitset>
#include <algorithm>
#include <stack>
#include <queue>
#include <map>
#include <climits>

using namespace std;

#define FOR(i, a, b) for(int (i)=a;(i)<(b);(i)++)
#define rep(i, n)  FOR(i,0,n)
#define ll long long
#define vi vector<int>
#define vvi vector<vector<int>>
#define vl vector<ll>
#define vvl vector<vector<ll>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
#define bigmod 1000000007
#define INF 1050000000


int main() {
    int N, A;
    cin >> N >> A;
    vi x(N);
    rep(i, N) cin >> x[i];
    int dp[N + 1][50 * N + 1][N + 1];
    rep(i, N + 1) rep(j, 50 * N + 1) rep(k, N + 1) dp[i][j][k] = 0;
    dp[0][0][0] = 1;
    //sort(x.begin(), x.end());
    rep(i, N) {
        rep(j, 50 * i + 1) {
            rep(k, i + 1) {
                dp[i + 1][j][k] += dp[i][j][k];
                dp[i + 1][j + x[i]][k + 1] += dp[i][j][k];
            }
        }
    }

    ll ans=0;
    FOR(i,1,N+1){
        ans += dp[N][A*i][i];
    }
    cout<<ans<<endl;


    return 0;
}