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: Time Limit Exceeded)

import java.util.*;

public class Main{
    public static void main(String[] args){
	Scanner sc = new Scanner(System.in);
	int N = sc.nextInt();
	int K = sc.nextInt();
	int[] A = new int[N+1];
	A[0] = 0;
	for (int i=1;i<=N;i++) {
	    A[i] = sc.nextInt();
	}
	for (int k=1;k<=K;k++) {
	    A = reA(A,N);
	}
	StringBuilder sb = new StringBuilder();
	for (int i=1;i<=N;i++) {
	    sb.append(A[i] + " ");
	}
	System.out.print(sb.toString());
	
    }
    static int[] reA(int[] A ,int N) {
	 int[] countL = new int[N+1];
	 for (int a=1;a<=N;a++) {
	     int str;
	     if (a-A[a] < 0) { str = 0;}
	     else { str = a-A[a]; }
	     int dst;
	     if (a+A[a] > N) { dst = N; }
	     else {dst = a+A[a]; }
	     
	     for (int ll=str; ll<=dst; ll++) {
		 countL[ll]++;
	     }
	 }
	 
	 return countL;
    }
}