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 Scala (2.11.7) (judge verdict: Time Limit Exceeded)

import java.io.{BufferedReader, InputStreamReader}

object Main extends App {
  val in = new BufferedReader(new InputStreamReader(System.in))
  val Array(h, w) = in.readLine().split(' ').map(_.toInt)
  val field = Array.fill(h)(in.readLine().toCharArray)
  val queue = new java.util.ArrayDeque[((Int, Int), Int)]()
  for (j <- 0 until h; i <- 0 until w; if field(j)(i) == '#') queue.offer(((i, j), 0))
  var times = 0
  while(!queue.isEmpty) {
    val ((x, y), count) = queue.poll()
    for {
      p <- Seq((x,y-1), (x,y+1), (x-1,y), (x+1,y))
      if p._1 >= 0 && p._1 < w && p._2 >= 0 && p._2 < h
      if field(p._2)(p._1) == '.'
    } {
      field(p._2)(p._1) = '#'
      queue.offer((p, count+1))
    }
    if (times < count) times = count
  }
  println(times)
}