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: Wrong Answer)

import java.util.*;
import java.io.*;
public class Main {
    static int arr[];
    static long dp[][],mod=1000000007L;
    public static void main(String args[]) {
        Scanner input=new Scanner(System.in);
        int n=input.nextInt();
        int k=input.nextInt();
        arr=new int[n];
        long sum=0;
        for(int i=0;i<n;i++) {
            arr[i]=input.nextInt();
            sum+=arr[i];
        }
        if(sum<k) {
            System.out.println(0);
            System.exit(0);
        }
        System.out.println(solve(n,k));
    }
    public static long solve(int n,int k) {
        dp=new long[n][k+1];
        for(int i=0;i<=arr[0];i++) {
            dp[0][i]=1;
        }
        for(int i=1;i<n;i++) {
            dp[i][0]=dp[i-1][0];
            for(int j=1;j<=arr[i];j++) {
                dp[i][j]=(dp[i][j-1]+dp[i-1][j])%mod;
            }
        }
        return dp[n-1][arr[n-1]];
    }
}