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 PHP7 (7.0.15) (judge verdict: Wrong Answer)

<?php
fscanf(STDIN, "%d %d", $H, $W);
$queue = [];

for ($i = 0; $i < $H; $i++) {
    $map[$i] = str_split(trim(fgets(STDIN)));

    for ($j = 0; $j < $W; $j++) {
        $visited[$i][$j] = -1;

        if ($map[$i][$j] == "#") {
            array_unshift($queue, [$i, $j]);
            $visited[$i][$j] = 0;
        }
    }
}

// BFS
$dx[0] = 0;
$dx[1] = 0;
$dx[2] = 1;
$dx[3] = -1;

$dy[0] = 1;
$dy[1] = -1;
$dy[2] = 0;
$dy[3] = 0;

while (count($queue) > 0) {
    $now_x = $queue[0][0];
    $now_y = $queue[0][1];
    array_shift($queue);

    for ($i = 0; $i < 4; $i++) {
        $next_x = $now_x + $dx[$i];
        $next_y = $now_y + $dy[$i];

        if ($next_x < 0 || $next_x >= $W || $next_y < 0 || $next_y >= $H) {
            continue;
        }

        if ($visited[$next_x][$next_y] == -1) {
             $queue[] = [$next_x, $next_y];
             $visited[$next_x][$next_y] = $visited[$now_x][$now_y] + 1;
        }
    }
}

$cnt = 0;

foreach ($visited as $k => $v) {
    foreach ($v as $kk => $vv) {
        if ($cnt < $vv) {
            $cnt = $vv;
        }
    }
}

echo $cnt."\n";