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 {
	public static void main (String[] args) {
      Scanner in = new Scanner(System.in);
      int N = in.nextInt();
      int K = in.nextInt();
      int[] a = new int[N+1];
      for (int i = 1; i <= N; i++) {
        a[i] = in.nextInt();
      }
      int[][] dp = new int[N+1][K+1];
      for(int i=0; i<= N ;i++){
        dp[i][0] = 1;
      }
      int[] temp = new int[K+1];
      int modulo = 1000000000+7;
      for(int i = 0; i<= N ;i++){
        temp[0] = 1;
        for(int j = 1; j<=K; j++){
          dp[i][j] = temp[j];
          
          if(i != N){
            temp[j] = (temp[j] % modulo + temp[j-1] % modulo) % modulo;
            temp[j] = (temp[j] % modulo - (( j-a[i+1]-1 >=0) ? dp[i][j-a[i+1]-1] % modulo : 0)) % modulo;
          }
        }
      }
      System.out.println(dp[N][K] % (modulo));
	}
}