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: Time Limit Exceeded)

class Point(val x: Int, val y: Int)

fun main(args: Array<String>) {
    val (h, w) = readLine()!!.split(" ").map { Integer.parseInt(it) }
    val count: MutableList<MutableList<Int>> = mutableListOf()
    repeat(h) {
        count.add(readLine()!!.toList().map { if (it == '#') 0 else -1 } as MutableList<Int>)
    }
    val queue: MutableList<Point> = mutableListOf()

    for (i in 0..h-1) {
        for (j in 0..w-1) {
            if (count[i][j] == 0) {
                queue.add(Point(i, j))
            }
        }
    }

    while (queue.size != 0) {
        val src = queue.removeAt(0)
        val dst = mutableListOf(
            Point(src.x - 1, src.y),
            Point(src.x + 1, src.y),
            Point(src.x, src.y - 1),
            Point(src.x, src.y + 1)
        ).filter {
            (0..h-1).contains(it.x) && (0..w-1).contains(it.y) && count[it.x][it.y] < 0
        }
        dst.forEach {
            count[it.x][it.y] = count[src.x][src.y] + 1
        }
        queue.addAll(dst)
    }
    println(count.flatten().max())
}