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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int A[1000][1000];

int main(void) {
    char buf[1001];
    int
        H, W, s,
        n = 0,
        time = 0;

    fgets(buf, 1001, stdin);
    H = atoi(strtok(buf, " "));
    W = atoi(strtok(NULL, " "));
    s = H * W;

    for (int i = 0; i < H; i++) {
        fgets(buf, 1001, stdin);
        for (int j = 0; j < W; j++) {
            if (buf[j] == '#') {
                A[i][j] = 0;
                n++;
            } else {
                A[i][j] = -1;
            }
        }
    }

    while (n < s) {
        for (int i = 0; i < H; i++) {
            for (int j = 0; j < W; j++) {
                if (A[i][j] == time) {
                    // 上
                    if (i > 0 && A[i-1][j] == -1) {
                        A[i-1][j] = time+1;
                        n++;
                    }
                    // 左
                    if (j > 0 && A[i][j-1] == -1) {
                        A[i][j-1] = time+1;
                        n++;
                    }
                    // 右
                    if (j < W-1 && A[i][j+1] == -1) {
                        A[i][j+1] = time+1;
                        n++;
                    }
                    // 下
                    if (i < H-1 && A[i+1][j] == -1) {
                        A[i+1][j] = time+1;
                        n++;
                    }
                }
            }
        }
        time++;
    }

    printf("%d\n", time);
    return 0;
}