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.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.*;
import java.lang.Math.*;
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        int n = in.nextInt();
        int k = in.nextInt();
        int [][] dp = new int[n+1][k+1];
        int []ai = new int[n+1];
        int mod = (int)1e9+7;
        dp[0][0]=1;
        for(int i=1;i<=n;i++)
            ai[i] = in.nextInt();
        for(int i=1;i<=n;i++){
            int []psum = new int[k+2];
            int cur =0;
            for(int idex=0;idex<=k;idex++){
                cur= (cur + dp[i-1][idex])%mod;
                psum[idex+1]=cur%mod;
            }
            for(int j=0;j<=k;j++){
                int mx = Math.min(ai[i],j);
                int a= psum[j+1];
                int b = psum[(j-mx)];
                if (a<b)
                    a+=mod;
                dp[i][j] = (a-b)%mod;
            }
        }
        System.out.println(dp[n][k]%mod);


    }
}
