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 <iostream>
#include <string>
#include <iterator>
#include <unordered_set>
#include <unordered_map>
#include <algorithm>
#include <sstream> 
#include <vector>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <iomanip>
#include <functional>
#pragma GCC optimize("O3")
using namespace std;
#define fast ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define ll long long
#define sp system("pause");
#define MOD 1000000007;
#define mxN 2005
#define MXINT 1e9+5
void solve()
{
	ll n, k;

	cin >> n >> k;
	vector<ll>a(n);
	for (int i = 0; i < n; i++)
	{
		cin >> a[i];
	}
	vector<vector<ll>> dp(n);
	for (int i = 0; i < n; i++)
		dp[i].resize(k+1);
	//k+1*n
	dp[0][0] = 1;
	for (int i = 1; i <= k; i++)
		dp[0][i] = dp[0][i - 1] + (a[0] >= i ? 1 : 0);
	for (int i = 0; i < n; i++)
		dp[i][0] = 1;
	for (int i = 1; i <= k; i++)
	{
		for (int j = 1; j < n-1; j++)
		{
			int m = min(a[j], k) + 1;
			dp[j][i] = (dp[j - 1][i] + dp[j][i - 1]) % MOD;
			if (i - m >= 0)
			{
			dp[j][i] -= dp[j - 1][i - m];
			}
		}
	}
	ll ans;
	if (n == 1)
	{
		if (a[0] < k)
			ans = 0;
		else
			ans = 1;
	}
	else
	{
		int m = min(a[n - 1], k) + 1;
		ans = dp[n - 2][k];
		if (k - m >= 0)
			ans -= dp[n - 2][k - m];

	}
	cout << ans % MOD;

}
int main()
{
	fast;
	solve();
	
	return 0;
}