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++14 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <vector>
const int MOD = 1e9+7;

int main() {
	int n,k;
	std::cin>>n;
	std::cin>>k;
	std::vector<int> a(n,0);
	int total = 0;
	for (int i = 0; i < n; ++i) {
		std::cin>>a[i];
		total += a[i];
	}
	if (k > total) {
		printf("0\n");
		return 0;
	}
	long long dp[n+1][k+1];
	long long s[k+2];
	memset(dp, 0, sizeof(dp));
	memset(s, 0, sizeof(s));
	for (int i = 0; i<=k; ++i)
		dp[0][i] = 0;
	for (int i = 0; i<=n; ++i)
		dp[i][0] = 1;
	for (int i = 1; i<=n; ++i) {
		s[0] = 0;
		for (int j = 1; j <= k+1; ++j)
			s[j] = s[j-1] + dp[i-1][j-1];
		for (int j = 1; j <=k; ++j) {
			dp[i][j] = s[j+1] - s[std::max(0, j - a[i-1])];
			dp[i][j] %= MOD;
		}
	}
	printf("%lld\n", dp[n][k]);

	return 0;
}