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)

#include <bits/stdc++.h>

#define PI 3.1415926535897932
#define MOD 1000000007
#define INF 1e9

#define REP(i, n) for(int i = 0; i < (n); i++)
#define REP1(i, n) for(int i = 1; i <= (n); i++)
#define ALL(v) (v).begin(), (v).end()
#define YES cout << "YES" << endl
#define Yes cout << "Yes" << endl
#define NO  cout << "NO"  << endl
#define No  cout << "No"  << endl

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef vector<bool> vb;

int dp[51][51][2501]; // i番目までのカードからj枚選んで合計がk

void solve()
{
    int N, A; cin >> N >> A;
    vi x(N+1); REP1(i, N) cin >> x[i];

    dp[0][0][0] = 1;

    REP(i, N+1)
    {
        REP(j, N+1)
        {
            REP(k, 2501)
            {
                if(i == 0 && j == 0 && k == 0) dp[i][j][k] = 1;
                else if(i >= 1 && x[i] > k) dp[i][j][k] = dp[i-1][j][k];
                else if(i >= 1 && j >= 1 && k >= x[i]) dp[i][j][k] = dp[i-1][j-1][k-x[i]] + dp[i-1][j][k]; 
                else dp[i][j][k] = 0;
            }
        }
    }
    
    ll ans = 0;
    REP1(j, N)
    {
        ans += dp[N][j][j*A];
    }
    cout << ans << endl;
}
    
int main() 
{
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);

    solve();

    return 0;
}