#include <bits/stdc++.h>
using namespace std;
const int N = 110, M = 10010;
//f[i][j]:考虑前i个数字,总和为j的所有方案
int f[N][M];
int a[N];
int n, m;
int main()
{
cin >> n >> m;
for(int i = 1; i <= n; i++) cin >> a[i];
f[0][0] = 1;
for(int i = 1; i <= n; i++)
{
for(int j = 0; j <= m; j++)
{
//不选
f[i][j] = f[i-1][j];
//选
if(j >= a[i])
f[i][j] += f[i-1][j-a[i]];
}
}
cout << f[n][m] << endl;
return 0;
}