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: Runtime Error)

#include "bits/stdc++.h"
#include "math.h"
 
using namespace std;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef vector<int> vin;
typedef pair<ll,ll> P;
typedef vector<P> vp;
#define rep(i,a,b) for(ll i=(a);i<(b);++i)
#define drep(i,a,b) for(ll i=(a);i>=(b);--i)
#define SIZE(a) ll((a).size())
#define out(a) cout<<(a)<<endl;
const int inf=INT_MAX;
const int MAX = 510000;
const ll MOD = 1000000007;

int main(){
  ll n,a;cin>>n>>a;
  vll x(n);
  rep(i,0,n)cin>>x[i];
  vector<vvll> dp(51,vvll(50,vll(2510,0))); 
  rep(i,0,n+1){
    dp[i][0][0]=1;
  }
  
  rep(i,1,n+1){
    rep(j,1,i+1){

      rep(k,0,2510){
        if(k<x[i-1]){dp[i][j][k]=dp[i-1][j][k];}
        else{dp[i][j][k]=dp[i-1][j][k]+dp[i-1][j-1][k-x[i-1]];}
      }
    }
  }
  ll ans=0;
  rep(i,1,n+1){
    ans+=dp[n][i][i*a];
  }
  out(ans);
}