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)

// M - Candies  https://atcoder.jp/contests/dp/tasks/dp_m
/* 入力例
3 4
1 2 3
// →出力 5
*/
#include <bits/stdc++.h>
using namespace std;

int main(void) {
    int N, K;
    cin >> N >> K;
    int a[N], dp[K + 1][N];
    for(int i = 0; i < N; i++) {
        cin >> a[i];
    }
    for(int j = 0; j < N; j++) {
        dp[0][j] = 1;
    }
    for(int i = 1; i <= K; i++) {
        if(i <= a[0]) {
            dp[i][0] = 1;
        } else {
            dp[i][0] = 0;
        }
        for(int j = 1; j < N; j++) {
            if(a[j-1] == 0) {
                dp[i][j] = dp[i][j - 1];
            } else {
                if(i <= a[j-1]) {
                    dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % 1000000007;
                } else {
                    dp[i][j] = (dp[i][j - 1] + dp[i - 1][j] - dp[i - 1 - a[j-1]][j - 1]) % 1000000007;
                }
            }
        }
    }
    cout << dp[K][N - 1] << endl;
    return 0;
}