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.Arrays;
import java.util.Scanner;

public class Main {

	static long mod = 1000_000_007;

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int N = in.nextInt(), K = in.nextInt();
		int a[] = new int[N + 1];
		int capa[] = new int[N + 1];
		for ( int i = 1; i <= N; i++ ) {
			a[i] = in.nextInt();

		}
		in.close();

		Arrays.sort(a);
		for ( int i = 1; i <= N; i++ ) {
			capa[i] += a[i] + capa[i - 1];
		}

		long dp[][] = new long[N + 1][K + 1];
		for ( int n = 0; n <= N; n++ ) {
			dp[n][0] = 1L;
			Arrays.fill(dp[n], 1, K + 1, 0);
		}

		for ( int n = 1; n <= N; n++ ) {
			if ( a[n] == 0 ) {
				dp[n] = dp[n - 1];
				continue;
			}
			for ( int k = 1; k <= K; k++ ) {

				if ( k <= a[n] ) {
					dp[n][k] += (dp[n][k - 1] + dp[n - 1][k]) % mod;
				} else /*if ( k <= a[n] )*/ {
					dp[n][k] = (dp[n][k - 1] + dp[n - 1][k] - dp[n - 1][k - a[n] - 1]) % mod;
				} /*else {
					break;
				}*/
			}
		}

		for ( long[] ls : dp ) {
			for ( long l : ls ) {
				System.out.print(l + " ");
			}
			System.out.println();
		}
		System.out.println(dp[N][K] % mod);
	}
}