Problem:
Score : 500 points

Problem Statement

We have N bulbs arranged on a number line, numbered 1 to N from left to right.
Bulb i is at coordinate i.

Each bulb has a non-negative integer parameter called intensity.
When there is a bulb of intensity d at coordinate x, the bulb illuminates the segment from coordinate x-d-0.5 to x+d+0.5.
Initially, the intensity of Bulb i is A_i. We will now do the following operation K times in a row:

For each integer i between 1 and N (inclusive), let B_i be the number of bulbs illuminating coordinate i. Then, change the intensity of each bulb i to B_i.

Find the intensity of each bulb after the K operations.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq K \leq 2 \times 10^5

0 \leq A_i \leq N

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \ldots A_N

Output

Print the intensity A{'}_i of each bulb i after the K operations to Standard Output in the following format:

A{'}_1 A{'}_2 \ldots A{'}_N

Sample Input 1

5 1
1 0 0 1 0

Sample Output 1

1 2 2 1 2

Initially, only Bulb 1 illuminates coordinate 1, so the intensity of Bulb 1 becomes 1 after the operation.
Similarly, the bulbs initially illuminating coordinate 2 are Bulb 1 and 2, so the intensity of Bulb 2 becomes 2.

Sample Input 2

5 2
1 0 0 1 0

Sample Output 2

3 3 4 4 3

Solution in Rust (1.42.0) (judge verdict: Time Limit Exceeded)

#![allow(unused)]

use proconio::input;
use proconio::marker::{Bytes, Chars, Usize1};
use std::cmp::{Reverse, min, max};
use itertools::Itertools;

macro_rules! debug {
    ($($a:expr),*) => {
        eprintln!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
    }
}

#[proconio::fastout]
fn main() {
    input! {
        n: usize,
        k: usize,
        mut xs: [i64; n],
    }

    let mut diffs = vec![0; n + 1];
    for i in 0..n {
        let xi = xs[i] as usize;
        diffs[if i < xi { 0 } else { i - xi }] += 1;
        diffs[min(i + xi + 1, n)] -= 1;
    }

    for _ in 0..k {
        let mut c = 0;
        let mut newdiffs = vec![0; n + 1];

        for i in 0..n {
            c += diffs[i];
            xs[i] = c;

            let xi = xs[i] as usize;
            newdiffs[if i < xi { 0 } else { i - xi }] += 1;
            newdiffs[min(i + xi + 1, n)] -= 1;
        }

        diffs = newdiffs;
    }

    println!("{}", xs.iter().join(" "));
}