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

const int N = 110, mod = 1000007;

int n, m;
int f[N];

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

    f[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 = m; j >= 0; j--)
            for (int k = 1; k <= min(a, j); k++)
                f[j] = (f[j] + f[j - k]) % mod;
    }

    cout << f[m] << endl;

    return 0;
}