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 <bits/stdc++.h>
using namespace std;
int n,m,k,a[109];
long long dp[109][100009],s[109][100009];
long long mod = 1000000007;
int main()
{
    cin>>n>>k;
    for(int i=1;i<=n;i++)cin>>a[i];
    for(int i=0;i<=k;i++) s[0][i] = 1;
    for(int i=1;i<=n;i++)
    {
        dp[i][0]=1;
        s[i][0]=1;
        for(int j=1;j<=k;j++)
        {
            dp[i][j]=(s[i-1][j]%mod);
            if(j-a[i]>0)dp[i][j]=((dp[i][j])%mod-(s[i-1][j-a[i]-1])%mod)%mod;
            s[i][j]=((dp[i][j])%mod+(s[i][j-1])%mod)%mod;
            dp[i][j]%=mod;
            s[i][j]%=mod;
        }
    }
    cout<<dp[n][k]%mod<<"\n";
    /*for(int i=0;i<=n;i++)
    {
        for(int j=0;j<=k;j++)cout<<dp[i][j]<<" ";
        cout<<"\n";
    }
    cout<<"_________\n";
    for(int i=0;i<=n;i++)
    {
        for(int j=0;j<=k;j++)cout<<s[i][j]<<" ";
        cout<<"\n";
    }*/
}