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 Scala (2.13.1) (judge verdict: Time Limit Exceeded)

object Main extends App {
  val inputs: Seq[String] = scala.io.Source.stdin.getLines().toList
  
  val nums = inputs(0).split(" ").map(_.toInt)
  val n = nums(0)
  val k = nums(1)
  val lamps = inputs(1).split(" ").map(_.toInt)


  println(solve(n, k, lamps))

  private def solve(n: Int, k: Int, lamps: Seq[Int]): String = {
    var currentPower = lamps
    var nextPower    = Seq.fill(n)(0)

    (1 to k).foreach { _ =>

      // 範囲初期化
      val range = scala.collection.mutable.ArrayBuffer.fill(n + 1)(0)
      // 要素を走査
      (0 until n).foreach { i =>
        val power = currentPower(i)

        // 開始、終了を挟み込み
        val left = Math.max(0, i - power)
        val right = Math.min(i + power + 1, n)

        range.update(left,  range(left) + 1)
        range.update(right, range(right) - 1)
      }
      // 累積和を計算
      range.zipWithIndex.foldLeft(0){ case (acc, (v, i)) =>
        val a = acc + v
        range.update(i, a)
        a
      }
      // 末端が余分なので削除して次の光として保持
      nextPower = range.toSeq.dropRight(1)

      // 全体が最高に光ってたら終了
      if (nextPower == currentPower) {
        return currentPower.mkString(" ")
      }

      // 次の光を保持
      currentPower = nextPower
    }
    currentPower.mkString(" ")
  }
}