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;

const int N = 110, K = 1e5 + 5;
const ll mod = 1e9 + 7;
ll n, k, a[N], dp[N][K], pref[K];

int main()
{
	ios_base::sync_with_stdio(false);

	cin >> n >> k;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	memset(dp, 0, sizeof(dp));
	dp[0][0] = 1;
	for (int i = 1; i <= n; i++)
	{
		pref[0] = dp[i - 1][0];
		for (int j = 1; j <= k; j++)
		{
			pref[j] = (pref[j - 1] + dp[i - 1][j]) % mod;
		}
		for (int j = 0; j <= k; j++)
		{
			ll give = pref[j], x = j - a[i] - 1;
			if (x >= 0)
				give -= pref[x];
			if (give < 0)
				give += mod;
			give = give % mod;
			dp[i][j] = give;
		}
	}
	cout << dp[n][k];

	return 0;
}