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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

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

public class Main {

	public static void main(String[] args) {
		Main ans = new Main();
		System.out.println(ans.solve());
	}

	private long solve() {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int K = sc.nextInt();
		int MOD = (int) 1e9+7;
		int [] array = new int[N+1];
		for(int i=1;i<=N;i++) {
			array[i] = sc.nextInt();
		}

		long[][] dp = new long[N+2][K+2];

		for(int i=0;i<=K;i++) {
			dp[1][i] = (i<=array[1]) ? 1 : 0;
		}


		for(int i=2;i<=N;i++){
			for(int j=0;j<=K;j++) {
				if (j==0) dp[i][j] = dp[i-1][j];
				else {
					long val = (j-1-array[i]) >= 0 ? dp[i-1][j-1-array[i]] : 0;		
					dp[i][j] = (MOD+ dp[i-1][j] + dp[i][j-1] - val)%MOD;	
				}				
			}
		}

		return dp[N][K];

	}

}