Problem:
Score : 100 points

Problem Statement

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.

Sample Input 1

3 4
1 2 3

Sample Output 1

5

There are five ways for the children to share candies, as follows:

(0, 1, 3)

(0, 2, 2)

(1, 0, 3)

(1, 1, 2)

(1, 2, 1)

Here, in each sequence, the i-th element represents the number of candies that Child i receives.

Sample Input 2

1 10
9

Sample Output 2

0

There may be no ways for the children to share candies.

Sample Input 3

2 0
0 0

Sample Output 3

1

There is one way for the children to share candies, as follows:

(0, 0)

Sample Input 4

4 100000
100000 100000 100000 100000

Sample Output 4

665683269

Be sure to print the answer modulo 10^9 + 7.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

/**
 *
 * @author Daipayan Mukherjee
 * @Jadavpur University,Information Technology
 *
 */
import java.io.*;
import java.util.*;

public class Main {
    public static long MOD = 1000000007;

    public static void main(String[] args)throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] st = br.readLine().split(" ");
        int n = Integer.parseInt(st[0]);
        int k = Integer.parseInt(st[1]);
        String[] s = br.readLine().split(" ");
        int[] cap = new int[n];
        for(int i=0;i<n;i++){
            cap[i] = Integer.parseInt(s[i]);
        }
        long[][] dp = new long[n][k+1];
        
        long last = 0;
        for(int i=0;i<=k;i++){
            long consumed =  k-i;
            if(consumed<=cap[0])
                dp[0][i] = (last+1)%MOD;
            else
                dp[0][i] = last;
            
            last = dp[0][i];
        }
   
        for(int i=1;i<n;i++){
            for (int j = 0; j <=k; j++) {
                long low = j>0?dp[i-1][j-1]:0;
                long high = dp[i-1][Math.min(k,j+cap[i])];
                dp[i][j] = ((j>0?dp[i][j-1]:0) + (high-low+MOD)%MOD)%MOD;
                
            }
        }
        System.out.println(dp[n-1][0]);
    }
}