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.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Hashtable;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

class Main
{
    long dp[][];
    int max[];
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int child=sc.nextInt();
        int candles=sc.nextInt();
        sc.nextLine();
        Main object=new Main();
        object.dp=new long[child+1][candles+1];
        object.max=new int[child];
        object.dp[0][0]=1;
        for(int i=1;i<=child;i++)
        {
          object.max[i-1]=sc.nextInt();
          for(int j=0;j<=candles;j++)
          {
              if(j-object.max[i-1]<=0)
                 object.dp[i][j]=object.dp[i-1][j]%1000000007;
              else
                 object.dp[i][j]=(object.dp[i-1][j]-object.dp[i-1][j-object.max[i-1]-1])%1000000007;
              if(j!=0)
                object.dp[i][j]=(object.dp[i][j]+object.dp[i][j-1]+1000000007)%1000000007;
          }
        }
        System.out.println(object.dp[child][candles]);
        sc.close();
     }
}