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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110;
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const int M = 10010;
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int a[N], f[N][M];
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int n, m;
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int main() {
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cin >> n >> m;
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for (int i = 1; i <= n; ++i)cin >> a[i];
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for (int i = 1; i <= n; ++i)
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for (int j = 1; j <= m; ++j) {
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//正好相等，方案数+1
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if (j == a[i])f[i][j] = f[i - 1][j] + 1;
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//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
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if (j > a[i]) f[i][j] = f[i - 1][j] + f[i - 1][j - a[i]];
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//如果小于，没的选择
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if (j < a[i]) f[i][j] = f[i - 1][j];
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}
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cout << f[n][m];
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return 0;
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}
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