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 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair <int, int> pii;
 
const int N = 1e5 + 10, mod = 1e9 + 7;

void add(int& a, int b) {
	a += b;
	if(a < 0) {
		a += mod;
	}
	if(a >= mod)
		a -= mod;
}

int dp[N];

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int n, k;
	cin >> n >> k;
	dp[0] = 1;
	for (int child = 0; child < n; child++) {
		int x;
		cin >> x;
		vector <int> fake(k + 1, 0);
		for (int used = k; used >= 0; used--) {
			int tmp = dp[used];
			int L = used + 1;
			int R = used + min(k - used, x);
			if(L <= R) {
				add(fake[L], tmp);
				if(R + 1 <= k) {
					add(fake[R + 1], -tmp);
				}
			}
		}
		int prefix_sum = 0;
		for (int i = 0; i <= k; i++) {
			add(prefix_sum, fake[i]);
			add(dp[i], prefix_sum);
		}
	}
	cout << dp[k] << endl;
}