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)

fun main(args: Array<String>) {
    val (n, k) = readLine()!!.split(' ').map { it.toInt() }
    var a = readLine()!!.split(' ').map { it.toInt() }.toIntArray()

    for (j in 0 until k) {
        val b = IntArray(n)
        for (i in 0 until n) {
            val llight = a[i]
            val max = Math.min(n-1, i+llight)
            for (l in i .. max) b[l]++

            val rlight = a[n-i-1]
            val min = Math.max(0, n-i-1-rlight)
            for (l in min .. n-i-2) b[l]++
        }
        a = b
    }
    println(a.joinToString(" "))
}