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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110, mod = 1000007;
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int n, m;
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int f[N][N]; // f[i, j] 表示前i个物品，总体积是j的方案数
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int main() {
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cin >> n >> m;
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f[0][0] = 1; // 求方案数的base case
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for (int i = 1; i <= n; i++) {
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int a;
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cin >> a; // 最多允许选a个
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for (int j = 0; j <= m; j++) {
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for (int k = 0; k <= min(a, j); k++)
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f[i][j] = (f[i][j] + f[i - 1][j - k]) % mod;
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}
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}
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cout << f[n][m] << endl;
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return 0;
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}
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