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: Time Limit Exceeded)

import java.io.*;
import java.util.*;

class Main {
    static final long MOD = 1000000007;

    // Generated by 1.1.7.1 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
    public static void main(String[] args) throws Exception {
        final Scanner sc = new Scanner(System.in);
        long N;
        N = sc.nextLong();
        long K;
        K = sc.nextLong();
        long[] a = new long[(int)(N)];
        for(int i = 0 ; i < N ; i++){
                a[i] = sc.nextLong();
        }
        solve(N, K, a);
    }

    static void solve(long N, long K, long[] a){
        long[][] dp = new long[(int)N][(int)(K+1)];
        
        for (int i = 0; i <= a[0]; i++){
            dp[0][i] = 1;
        }

        for (int i = 1; i < N; i++){
            for (int j = 0; j <= K; j++){
                long sum = 0;
                for (int k = 0; k <= a[i]; k++){
                    if (j >= k){
                        sum = (sum + dp[i-1][j-k]) % MOD;
                    }
                }
                dp[i][j] = sum;
            }
        }

        System.out.print(dp[(int)(N-1)][(int)(K)]);
    }
}