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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
map<long long,map<long long, long long> > f;
long long count(long long b,long long n){
	if(f[b][n]!=0){
		return f[b][n];
	}
	if(n<b){
		//cout << b << " " << n<<endl;
		return n;
	}
	else{
		f[b][n] = count(b,n/b) + n%b;
		return f[b][n];
	}
}
signed main(){
	long long n,s;
	cin >> n >> s;
	long long l=1;
	long long r=1e11;
	while(l!=r){
		long long mid= (l+r)/2;
		long long val=count(mid,n);
		if(val==s){
			cout << mid;
			return 0;
		}
		else if(val>s){
			r=mid;
		}
		else{
			l=mid+1;
		}
	}
	cout << "-1";
}