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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define int long long
#define p (int)(1e9 + 7)

int MOD(int a){
	a %= p;
	return a >= 0 ? a : a + p;
}

signed main(){
	int N, K, i, j;
	scanf("%lld%lld", &N, &K);
	int *a = (int *)malloc(sizeof(int) * (N + 1));
	for(i = 1; i <= N; i++){
		scanf("%lld", &a[i]);
	}
	int **dp = (int **)malloc(sizeof(int *) * (N + 1));
	int **s = (int **)malloc(sizeof(int *) * (N + 1));
	for(i = 0; i <= N; i++){
		dp[i] = (int *)malloc(sizeof(int) * (K + 1));
		s[i] = (int *)malloc(sizeof(int) * (K + 1));
		for(j = 0; j <= K; j++){
			dp[i][j] = 0;
			s[i][j] = 0;
		}
	}
	dp[0][0] = 1;
	for(j = 0; j <= K; j++){
		s[0][j] = 1;
	}
	for(i = 1; i <= N; i++){
		for(j = 0; j <= K; j++){
			if(j <= a[i]){
				dp[i][j] = s[i - 1][j];
			}
			else{
				dp[i][j] = MOD(s[i - 1][j] - s[i - 1][j - a[i] - 1]);
			}
		}
		s[i][0] = dp[i][0];
		for(j = 1; j <= K; j++){
			s[i][j] = MOD(s[i][j - 1] + dp[i][j]);
		}
	}
	printf("%lld\n", dp[N][K]);
	return 0;
}