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.*;
import java.io.*;

public class Main {

	public static final int mod = 1000000007;

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] s = br.readLine().split(" ");
		int n = Integer.parseInt(s[0]);
		int k = Integer.parseInt(s[1]);
		int[] arr = new int[n + 1];
		int[][] dp = new int[n + 1][k + 1];
		int[] prefix = new int[k+1];
		s = br.readLine().split(" ");

		dp[0][0] = 1;
		for (int i = 1; i <= n; i++) {
			arr[i] = Integer.parseInt(s[i - 1]);
			dp[i][0] = 1;
		}

		int max = 0;
		for (int i = 1; i <= n; i++) {
			prefix[0] = dp[i-1][0];
			max += arr[i];
//			System.out.println(max);
			for (int j = 1; j <= max && j <=k; j++) {
				prefix[j] = (dp[i-1][j]+ prefix[j-1]) % mod;
//				System.out.println(i + " " + j+ " " + prefix[j]);
				if (j-arr[i]-1 < 0) {
					dp[i][j] = prefix[j];
				} else {
					dp[i][j] = (prefix[j]-prefix[j-arr[i]-1]+mod)%mod;
				}
			}
		}
		System.out.println(dp[n][k]);
//		for (int j = 0; j <= k; j++) {
//			for (int i = 0; i <= n; i++) {
//				System.out.print(dp[i][j] + " ");
//			}
//			System.out.println();
//		}

	}

}
