#include <bits/stdc++.h>
using namespace std;

const int N = 110, mod = 1000007;

int n, m;
int f[N][N]; // f[i, j] 表示前i个物品，总体积是j的方案数

int main() {
    cin >> n >> m;

    f[0][0] = 1; // 求方案数的base case

    for (int i = 1; i <= n; i++) {
        int a;
        cin >> a; // 最多允许选a个
        for (int j = 0; j <= m; j++) {
            for (int k = 0; k <= min(a, j); k++)
                f[i][j] = (f[i][j] + f[i - 1][j - k]) % mod;
        }
    }

    cout << f[n][m] << endl;

    return 0;
}