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)

import java.util.Scanner;

public class Main{
	static long mod = 1000000000 + 7;
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		while(sc.hasNext()) {
			int N = sc.nextInt(), K = sc.nextInt();
			int[] a = new int[N];
			for(int i = 0; i < N; i++) a[i] = sc.nextInt();
			long[][] dp = new long[N][K + 1];
			for(int i = 0; i <= a[N - 1]; i++) dp[N - 1][i] = 1;
			for(int i = N - 2; i >= 0; i--) {
				for(int j = 0; j <= K; j++) {
					if(j == 0) {
						dp[i][j] = dp[i + 1][j];
						continue;
					}
					dp[i][j] = dp[i + 1][j] + dp[i][j - 1];
					if(j - a[i] - 1 >= 0) {
						dp[i][j] -= dp[i + 1][j - a[i] - 1];
						dp[i][j] = (mod + dp[i][j]) % mod;
					}
					dp[i][j] %= mod;
				}
			}
			System.out.println(dp[0][K] % mod);
		}
	}
}