#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;
}