There are N children, numbered 1, 2, \ldots, N.

They have decided to share K candies among themselves. Here, for each i (1 \leq i \leq N), Child i must receive between 0 and a_i candies (inclusive). Also, no candies should be left over.

Find the number of ways for them to share candies, modulo 10^9 + 7. Here, two ways are said to be different when there exists a child who receives a different number of candies.

Constraints

* All values in input are integers.
* 1 \leq N \leq 100
* 0 \leq K \leq 10^5
* 0 \leq a_i \leq K

Input

Input is given from Standard Input in the following format:


N K
a_1 a_2 \ldots a_N


Output

Print the number of ways for the children to share candies, modulo 10^9 + 7.

Examples

Input

3 4
1 2 3


Output

5


Input

1 10
9


Output

0


Input

2 0
0 0


Output

1


Input

4 100000
100000 100000 100000 100000


Output

665683269
import java.util.*;

public class Main {
    static long[][] dp;
    static final long MOD = 1_000_000_007;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int K = sc.nextInt();
        int[] a = new int[N];
        for (int i = 0; i < N; i++) {
            a[i] = sc.nextInt();
        }
        dp = new long[N + 1][K + 1];
        dp[0][0] = 1;
        for (int k = 1; k < K + 1; k++) {
            dp[0][k] = 0;
        }
        for (int i = 0; i < N; i++) {
            long[] diff = new long[K + 1];
            for(int k = 0; k < K + 1; k++) diff[k] = 0;
            for (int k = 0; k < K + 1; k++) {
                diff[k] += dp[i][k];
                diff[k] = properMod(diff[k]);
                int end = k + a[i] + 1;
                if (end < K + 1) {
                    diff[end] -= dp[i][k];
                    diff[end] = properMod(diff[end]);
                }
            }
            dp[i + 1][0] = diff[0];
            for (int k = 1; k < K + 1; k++) {
                dp[i + 1][k] = dp[i + 1][k - 1] + diff[k];
                dp[i + 1][k] = properMod(dp[i + 1][k]);
            }
        }
        long ans = dp[N][K];
        System.out.println(ans);
    }
    public static long properMod(long n) {
        if (n >= 0) {
            return n%MOD;
        } else {
            return (MOD + n%MOD);
        }
    }
}