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#include <bits/stdc++.h>
using namespace std;
const int N = 110;
const int M = 10010;
int a[N], f[N][M];
int n, m;
int main() {
cin >> n >> m;
for (int i = 1; i <= n; ++i)cin >> a[i];
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= m; ++j) {
//正好相等，方案数+1
if (j == a[i])f[i][j] = f[i - 1][j] + 1;
//如果大于，不选，就是原来的方案数；选了，那么就依赖于j-a[i]这些钱在i-1个物品中的方案数
if (j > a[i]) f[i][j] = f[i - 1][j] + f[i - 1][j - a[i]];
//如果小于，没的选择
if (j < a[i]) f[i][j] = f[i - 1][j];
}
cout << f[n][m];
return 0;
