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)

#include<bits/stdc++.h>
using namespace std;
int main() 
{
    int n,k,i,j;
    cin>>n>>k;
    int m=1e9+7;
    int a[n+1],s[k+1];
    for(i=1;i<=n;i++)
    cin>>a[i];
    int dp[n+1][k+1];
    dp[0][0]=1;
    for(i=1;i<=k;i++)
    dp[0][i]=0;
    for(i=1;i<=n;i++)
    {
        s[0]=dp[i-1][0];
        for(j=1;j<=k;j++)
        s[j]=(s[j-1]+dp[i-1][j])%m;
        
        for(j=0;j<=k;j++)
        {
            int x=s[j];
            int y=j-a[i]-1;
            if(y>=0)
            x=(x-s[y])%m;
            dp[i][j]=x;
        }
    }
    cout<<dp[n][k];
  
}