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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

//Abraao Caiana de Freitas - UFCG

#include <bits/stdc++.h>
#define mod 1000000007
#define maxx 100010
using namespace std;
int dp[200][maxx],arr[200];

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n,m;
	cin>>n>>m;
	for(int i = 1; i <= n; i++){cin>>arr[i];dp[i][0]=1;}
	for(int i = 0; i <= m; i++)dp[1][i]=(i<=arr[1]?1:0);
	for(int i = 2; i <= n; i ++){
		for(int j = 1; j <= m; j++){
			int extra=0;
			if(j>arr[i])extra=dp[i-1][j-arr[i]-1];
			dp[i][j]=((mod+dp[i-1][j]+dp[i][j-1])-extra)%mod;
		}
	}
	cout<<dp[n][m]<<"\n";
	return 0;
}