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)

use proconio::input;

const MOD: i64 = 1000000007;

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

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