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.h>
#include <bits/stdc++.h>
#include<algorithm> 
using namespace std;

int main() {
int N, K;
cin >> N >> K;
int a[N];

for (int i = 0; i < N; ++i)
{
	cin >> a[i];
}

if(N == 1){
	if(a[0] < K) cout << 0 << endl;
	else cout << 1 << endl;
	return 0;
}

long int dp[N][K+1];

for (int i = 0; i < N; ++i)
{
	for (int j = 0; j <= K; ++j)
	{
		dp[i][j] = 0;
	}
}

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

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

for (int i = 1; i < N; ++i)
{
	for (int j = 0; j <= K; ++j)
	{	
		if(j <= a[i])
		dp[i][j] = (j > 0) ? (dp[i][j-1] + dp[i-1][j])%1000000007 : dp[i-1][j]%1000000007;
		else dp[i][j] = dp[i][j] + dp[i][j-1];
	}
}
if(a[N-1] != 0)
cout << (dp[N-1][a[N-1]] - dp[N-1][a[N-1] - 1] + 1000000007)%1000000007<< endl;
else cout << dp[N-1][a[N-1]]%1000000007 << endl;

return 0;
                                                     
}