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)

// Author : Syed Mujtaba
// Seek knowledge from the Cradle to the Grave !!
// https://www.linkedin.com/in/syed-mujtaba-064a5016a 

#include <bits/stdc++.h>
using namespace std;
#define int     long long
#define fast_io ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mod ((int)(1e9 + 7))
signed main()
{
	fast_io
	int n, k;
	cin >> n >> k;
	int a[n + 1];
	int dp[n + 1][k + 1];
	vector < int > psa(n + 1, 0);
	for(int i = 1; i <= n; i++)
		cin >> a[i];
	for(int i = 1; i <= n; i++)
	{
		psa[i] = a[i] + psa[i - 1];
	}
	vector < int > ps[n + 4];
	for(int i = 0; i <= n; i++)
	{
		ps[i].resize(k + 4);
		for(int j = 0; j <= k; j++)
			ps[i][j] = 0;
	}
	for(int i = 0; i <= k; i++)
	{
		if(i <= a[1])
			dp[1][i] = 1;
		else
			dp[1][i] = 0;
		ps[1][i+1] = (ps[1][i]%mod + dp[1][i]%mod)%mod;
	}
	for(int i = 2; i <= n; i++)
	{
		dp[i][0] = 1;
		ps[i][1] = 1;
		for(int j = 1; j <= k; j++)
		{
			dp[i][j] = 0;
			dp[i][j] = (ps[i - 1][j + 1]%mod - ps[i - 1][max(0LL, j - a[i])]%mod)%mod;
			ps[i][j + 1] = (ps[i][j]%mod + dp[i][j]%mod)%mod;
		}
	}
	cout << dp[n][k] << endl;
}