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;
#define MOD 1000000007
#define ll long long

ll dp[105][100100];
int ar[105];

inline ll add(ll a, ll b)
{
	a = ((a % MOD) + MOD) % MOD;
	b = ((b % MOD) + MOD) % MOD;
	return (a + b) % MOD;
}

inline ll sub(ll a, ll b)
{
	a = ((a % MOD) + MOD) % MOD;
	b = ((b % MOD) + MOD) % MOD;
	return ((a - b) % MOD + MOD) % MOD;
}

int main()
{

	int n, k;
	scanf("%d", &n); scanf("%d", &k);

	for (int i = 0; i < n; ++i)
	{
		scanf("%d", &ar[i]);
	}

	for (int i = 0; i <= ar[0]; ++i)
	{
		dp[0][i] = 1;
	}

	for (int i = 1; i < n; ++i)
	{
		dp[i][0] = 1;
		for (int j = 1; j <= k; ++j)
		{
			dp[i][j] = add(dp[i][j - 1] , dp[i - 1][j]);
			if (j > ar[i])
				dp[i][j] = sub(dp[i][j], dp[i - 1][j - 1 - ar[i]]);
		}
	}

	cout << dp[n - 1][k];




	return 0;
}