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(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
 
    int n, k;
    cin>>n>>k;
 
    vector<int> A(n);
    for(int i=0;i<n;i++){
        cin>>A[i];
    }
 
//     9
// 100000
// 0 0 100000 100000 100000 100000 0 0 0 
// 5
// 100000
//  100000 100000 100000 100000 0
    vector<vector<int>> dp(2, vector<int>(k+1));
 
    int MOD=1e9+7;
 
    for(int j=0;j<=min(k, A[0]);j++){
        dp[0][j]=1;
    }
 
    for(int i=1;i<n;i++){
        int current=i%2;
        int last=!current;
        dp[current][0]=1;
        long long temp=1;
        for(int j=1;j<=A[i];j++){
            temp=(temp+(long long)dp[last][j])%(long long)MOD;
            dp[current][j]=temp;
        }
 
        for(int j=A[i]+1; j<=k; j++){
            if(j-A[i]-1>=0){
                temp=(temp-(long long)dp[last][j-A[i]-1]+MOD)%(long long)MOD;
            }
            long long temp1=(temp+(long long)dp[last][j]+MOD)%(long long)MOD;
            dp[current][j]=temp1;
        }
    }
 
    int current=(n-1)%2;
    cout<<dp[current][k]<<endl;
}