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.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int K = sc.nextInt();

		// A:値を保存する主配列
		// B:累積和を表現する副配列
		int[] A = new int[N + 1];
		int[] B = new int[N + 1];

		// Aの要素に入力を代入
		for (int i = 0; i < N; i++) {
			int a = sc.nextInt();
			A[i] = a;
		}
		
		// K回の試行を行う主反復
		for (int i = 0; i < Math.min(K, 42); i++) {
			/* 解法
			 * 長さNの配列について累積和を計算。それぞれの加算範囲はAのn番目の要素ajについてn-aj~n+ajの範囲
			 * 配列外参照に注意
			 */
			// Bを初期化
			Arrays.fill(B, 0);
			for (int j = 0; j < N; j++) {
				// j番目の要素について、それぞれの始点、終点を記録。配列範囲より大きければ0もしくはNを指定
				// j - ajを始点とすると
				int start = Math.max(0, j - A[j]);
				// j + aj - 1が終点となる(終わりをふくむため)
				int end = Math.min(N, j + A[j] + 1);
				//始点を+1、終点を-1
				B[start]++;
				B[end]--;
			}
			for (int j = 1; j < N; j++) {
				// Bを順に加算していく
				B[j] += B[j - 1];
			}
			// Bの値をAとする
			A = B.clone();
		}

		// 出力
		for (int i = 0; i < N; i++) {
			System.out.print(A[i] + (i < N - 1 ? " " : "\n"));
		}
	}
}