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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

const int MAX = 1e5 + 5;
const int MOD = 1e9 + 7;

#define dbg(a) cout << "-> " << __LINE__ << ": " << #a << " = " << a << endl

int add (int a, int b) {
	return (a + b < MOD) ? (a + b) : (a + b - MOD);
}

int sub (int a, int b) {
	return (a > b) ? (a - b) : (a - b + MOD);
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n, k;
	cin >> n >> k;
	vector <int> a(n + 1);
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	vector <int> dp(k + 1, 1), cp(k + 1, 0);
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j <= k; j++) {
			int l = max(j - a[i], 0), r = j;
			cp[j] = (l > 0) ? sub(dp[r], dp[l - 1]) : dp[r];
			if (j > 0) {
				cp[j] = add(cp[j], cp[j - 1]);
			}
		}
		dp = cp;
	}
	dp[k] = sub(dp[k], ((k > 0) ? dp[k - 1] : 0));
	cout << dp[k] << '\n';
	return 0;
}