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 Kotlin (1.3.71) (judge verdict: Time Limit Exceeded)

import java.io.*
import java.util.*
import kotlin.math.*

fun main(){
	val f = BufferedReader(InputStreamReader(System.`in`))

	val (n,m) = f.readLine().split(" ").map{it.toInt()}
	var prev = IntArray(1) + f.readLine().split(" ").map{it.toInt()}.toIntArray()
	var cur = IntArray(n+1){1}

	var i = 1
	while(i <= m){
		val pqr = PriorityQueue<Int>()

		for(k in 1..n){
			while(!pqr.isEmpty() && pqr.peek() < k) pqr.poll()
			cur[k] += pqr.size
			pqr.add(k+prev[k])
		}

		val pql = PriorityQueue<Int>(compareBy{it*-1})
		for(k in n downTo 1){
			while(!pql.isEmpty() && pql.peek() > k) pql.poll()
			cur[k] += pql.size
			pql.add(k-prev[k])
		}

		var done = true
		for(k in 1..n){
			prev[k] = cur[k]
			cur[k] = 1
			if(prev[k] < n) done = false
		}

		if(done) break
		i++
	}

	println(prev.joinToString(" ").substring(2))

}