Problem:
Score : 300 points

Problem Statement

You are given a grid of squares with H horizontal rows and W vertical columns, where each square is painted white or black.
HW characters from A_{11} to A_{HW} represent the colors of the squares.
A_{ij} is # if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is . if that square is white.

We will repeatedly perform the following operation until all the squares are black:

Every white square that shares a side with a black square, becomes black.

Find the number of operations that will be performed.
The initial grid has at least one black square.

Constraints

1 \leq H,W \leq 1000

A_{ij} is # or ..

The given grid has at least one black square.

Input

Input is given from Standard Input in the following format:

H W
A_{11}A_{12}...A_{1W}
:
A_{H1}A_{H2}...A_{HW}

Output

Print the number of operations that will be performed.

Sample Input 1

3 3
...
.#.
...

Sample Output 1

2

After one operation, all but the corners of the grid will be black. After one more operation, all the squares will be black.

Sample Input 2

6 6
..#..#
......
#..#..
......
.#....
....#.

Sample Output 2

3

Solution in Kotlin (1.0.0) (judge verdict: Runtime Error)

import java.util.*

fun main(args: Array<String>) {
    val (MAX_Y, MAX_X) = readLine()!!.split(" ").map(String::toInt)
    val array = Array(MAX_Y) {
        readLine()!!.toCharArray()
    }

    val INF = Integer.MAX_VALUE

    val dx = intArrayOf(1, 0, -1, 0)
    val dy = intArrayOf(0, 1, 0, -1)


    var max = 0
    fun bfs(): Int {
        val queue: Queue<Pair<Int, Int>> = ArrayDeque()

        val d = Array(MAX_Y) {
            IntArray(MAX_X) { INF }
        }

        for (x in 0 until MAX_Y) {
            for (y in 0 until MAX_X) {
                if (array[y][x] == '#') {
                    queue.add(x to y)
                    d[y][x] = 0
                }
            }
        }

        while (queue.isNotEmpty()) {
            val (x, y) = queue.poll()

            for (i in 0..3) {
                val nx = x + dx[i]
                val ny = y + dy[i]

                if (nx in 0 until MAX_X && ny in 0 until MAX_Y && d[ny][nx] == INF) {
                    queue.add(nx to ny)
                    d[ny][nx] = d[y][x] + 1
//                    array[ny][nx] = '#'
                    max = Math.max(max, d[ny][nx])
                }
            }

//            println("------------------")
//            array.joinToString(System.lineSeparator()) { it.joinToString("") }.let(::println)
//            println("------------------")

        }
        return max
    }

    println(bfs())
}