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: Wrong Answer)

// C++ 14
#include <bits/stdc++.h>
using namespace std;
template <typename T>  ostream& operator<<(ostream& os, const vector<T>& v)  {  os << "[";  for (int i = 0; i < v.size(); ++i) {  os << v[i];  if (i != v.size() - 1)  os << ", ";  }  os << "]";  return os;  }
template <typename T> void print(T v, string s = "\n") { cout << v << s; }
template <typename T> void in(T &v) { cin >> v; }
#define ll long long
#define loop(__x, __start, __end) for(int __x = __start; __x < __end; __x++)

int nth(ll S, int n) {
  return (S>>n)&1;
}
int main() {
  int n, k; in(n),in(k);
  vector<int> A(n);
  loop(i,0,n) in(A[i]);
  
  ll ans = 0;
  for (ll S=1; S<(1<<n); S++) {
    int sum1 = k * __builtin_popcount(S);
    int sum2 = 0;
    loop(i,0,n) {
      if (nth(S, i)) sum2 += A[i];
    }
    ans += sum1 == sum2;
  }
  print(ans);
  return 0;
}