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: Accepted)

// Created by sz
#include <bits/stdc++.h>
using namespace std;

const int maxn = 105, maxk = 1e5+5;
const int mod = 1e9+7;
int N, K, a[maxn], dp[maxn][maxk];

int main(){
#ifdef LOCAL
    freopen("./input.txt", "r", stdin);
#endif
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    

    cin>>N>>K;
    for(int i = 0; i<N; i++){
        cin>>a[i];
    }
    
    dp[0][0] = 1;
    for(int i = 0; i< N; i++){
        dp[i+1][0] = 1;
        for(int k = 1; k <= K; k++){
            dp[i+1][k] = ((dp[i+1][k-1]%mod + dp[i][k]%mod)%mod - (k-1>=a[i]?dp[i][k-1-a[i]]:0)%mod +mod)%mod;
//            for(int m = 0; m<=a[i]&&m<=k; m++){
//                dp[i+1][k] += (dp[i][k-m])%mod;
//                dp[i+1][k] %= mod;
//            }
        }
    }
    
    cout<<dp[N][K]<<endl;


    
    return 0;
}