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::vec::Vec;
use std::cmp::*;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().split_whitespace()
        .map(|e| e.parse().ok().unwrap()).collect()
}

const MOD: u64 = 1_000_000_000 + 7;

fn main() {
    let (n, k) = {
      let v: Vec<usize> = read_vec();
      (v[0], v[1])
    };

    let a: Vec<usize> = read_vec();
    // dp_table
    let mut dp: Vec<Vec<u64>> = vec![vec![0; k+1]; n+1];

    // init
    dp[0][0] = 1;

    for i in 0..n {
      // i列目を累積和に更新
      for j in 0..k {
        dp[i][j+1] += dp[i][j];
        dp[i][j+1] %= MOD;
      }
      //
      for j in 0..k+1 {
        // dp[i+1][j+1] = SUM(dp[i][l]) (for l = j - a_i ~ j)
        let l = max(0, j as i64 - a[i] as i64);

        if l == 0 {
          dp[i+1][j] = dp[i][j];
        } else  {
          dp[i+1][j] = dp[i][j] - dp[i][l as usize - 1];
        }

        dp[i+1][j] %= MOD;
      }
    }

    println!("{}", dp[n][k]);
}