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 (1.42.0) (judge verdict: Accepted)

#[allow(unused_imports)]
use std::cmp::Reverse;

#[allow(unused_imports)]
use proconio::{fastout, input, marker::Chars};

#[allow(unused_imports)]
use itertools::Itertools;

#[fastout]
pub fn main() {
    input! {
        n: usize,
        k: usize,
        a: [usize; n],
    }
    const MOD: usize = 1000000007;

    let mut dp = vec![vec![0; k + 1]; n + 1];
    dp[0][0] = 1;
    for i in 1..=n {
        let s: Vec<_> = dp[i - 1]
            .iter()
            .scan(0, |acc, x| {
                *acc += *x % MOD;
                Some(*acc)
            })
            .collect();
        for j in 0..=k {
            dp[i][j] = (s[j] + MOD - if j > a[i - 1] { s[j - a[i - 1] - 1] } else { 0 }) % MOD;
        }
    }
    println!("{}", dp[n][k]);
}