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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::*;
const MOD: usize = 1_000_000_007;

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let k: usize = itr.next().unwrap().parse().unwrap();
    let a: Vec<usize> = (0..n)
        .map(|_| itr.next().unwrap().parse().unwrap())
        .collect();

    let mut dp: Vec<Vec<usize>> = vec![vec![0; k + 1]; n + 1];
    dp[0][0] = 1;
    for i in 1..n + 1 {
        for j in 1..k + 1 {
            dp[i - 1][j] += dp[i - 1][j - 1];
            dp[i - 1][j] %= MOD;
        }
        for j in 0..k + 1 {
            if j > a[i - 1] {
                dp[i][j] += dp[i - 1][j] - dp[i - 1][j - a[i - 1] - 1];
            } else {
                dp[i][j] += dp[i - 1][j];
            }
            dp[i][j] %= MOD;
        }
    }
    println!("{}", dp[n][k]);
}