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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        
        int[] array = new int[n];
        for (int i = 0; i < n; i++) {
            array[i] = sc.nextInt();
        }
        
        // process: 「光が届く電球の数を光の強さにする」をk回繰り返す
        // ex. 1,0,0,1,0
        // *-
        //  *
        //   *
        //   -*-
        //     -    => 1,2,2,1,2 (k = 1)
        // *-
        // -*--
        // --*--
        //   -*-
        //   --*    => 3,3,4,4,3 (k = 2)
        
        while (k-- > 0) {
            int[] start = new int[n];
            int[] end = new int[n];
            for(int i = 0; i < n; i++) {
                start[i] = Math.max(0, i-array[i]);
                end[i] = Math.min(n-1, i+array[i]);
            }
            
            // imos法: 累積和の区間版
            // sum[i]: 時刻iで放送中の番組数
            int[] sum = new int[n+1];
            for (int i = 0; i < n; i++) {
                // 時刻start[i]で放送番組が1増える
                // 時刻end[i]+1で放送番組が1減る
                sum[start[i]]++;
                sum[end[i] + 1]--;
            }
            for (int i = 0; i < n; i++) {
                sum[i + 1] += sum[i];
            }
            
            long now = 0;
            for (int i = 0; i < n; i++) {
                array[i] = sum[i];
                now += array[i];
            }
            long max = (long)n * n;
            if (now == max) {
                for (int i = 0; i < n; i++) {
                    System.out.print(array[i]);
                    if (i != n - 1) {
                        System.out.print(" ");
                    } else {
                        System.out.println();
                    }
                }
                // System.out.println(k);
                return;
            }
            
            // System.out.println(Arrays.toString(sum));
        }
        
        for (int i = 0; i < n; i++) {
            System.out.print(array[i]);
            if (i != n - 1) {
                System.out.print(" ");
            } else {
                System.out.println();
            }
        }
    }
}