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;
#define int long long
const int MAXN = 1e2 + 5;
const int MAXK = 1e5 + 5;
const int MOD = 1e9 + 7;
int arr[MAXN];
int dp[MAXN][MAXK];
int sum[MAXN][MAXK];
int32_t main(){
    int n, k;
    cin >> n >> k;
    for(int i = 0; i < n; i++)
        cin >> arr[i];
    for(int i = 0; i < n; i++){
        dp[i][0] = 1;
        sum[i][0] = 1;
    }
    for(int j = 1; j <= k; j++){
        if(j <= arr[0]) dp[0][j] = 1;
        sum[0][j] = dp[0][j] + sum[0][j - 1];
    }
    for(int i = 1; i < n; i++){
        for(int j = 1; j <= k; j++){
            if(j - arr[i] - 1 >= 0)
                dp[i][j] = (sum[i - 1][j] - sum[i - 1][j - arr[i] - 1]) % MOD;
            else
                dp[i][j] = (sum[i - 1][j])%MOD;
        }
        for(int j = 1; j <= k; j++){
            sum[i][j] = (dp[i][j] + sum[i][j - 1])%MOD;
        }
    }
    cout << dp[n - 1][k] << '\n';
}