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 9.2.1) (judge verdict: Wrong Answer)

#include<iostream>

using namespace std;

#define ll long long
ll mod = 1000000007;
ll solve(ll v[], ll n, ll k) {
	ll dp[n + 2][k + 2];
	for (ll j = 0; j <= k; j++) {
		dp[1][j] = (j > v[1]) ? 0 : 1;
	}

	for (ll i = 2; i <= n; i++) {
		for (ll j = 0; j <= k; j++) {
			if (j == 0)
				dp[i][j] = dp[i - 1][j];
			else
				dp[i][j] = (mod + dp[i - 1][j] + dp[i][j - 1] - ((j - v[i] - 1 >= 0) ? dp[i - 1][j - v[i] - 1] : 0)) % mod;
		}
	}
	for (ll i = 1; i <= n; i++) {
		for (ll j = 0; j <= k; j++)
		{
			cout << dp[i][j] << " ";
		}
		cout << "\n";
	}
	return dp[n][k];



}
int main() {
	ll n, k;
	cin >> n >> k;
	ll v[n + 1];
	for (ll i = 1; i <= n; i++) {
		cin >> v[i];
	}
	cout << solve(v, n, k) << endl;

	return 0;
}