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>

#define _max(_x, _y) ((_x)>(_y)?(_x):(_y))
#define _min(_x, _y) ((_x)>(_y)?(_y):(_x))
#define _abs(_x) ((_x)>0?(_x):(-(_x)))

char A[1000][1000][1000+1];

int dump_A(int age, int H, int W) {
int i, j;
  printf("age(%d)\n", age);
  for (i = 0; i < H; ++i) {
    for (j = 0; j < W; ++j) {
      printf("%c ", A[age][i][j]);
    }
    printf("\n");
  }
}

int is_all_black(int age, int H, int W) {
int i, j;
  //dump_A(age, H, W);
  for (i = 0; i < H; ++i) {
    for (j = 0; j < W; ++j) {
      if (A[age][i][j] == '.') return 0;
    }
  }
  //printf("all_black!\n");
  return 1;
}


int near_black(int age, int i, int j, int H, int W) {
  ////////////////////////////////////////
  // １個上の行を見る
  ////////////////////////////////////////
  if (i > 0) {
    // 真上を見る
    if (A[age][i-1][j] == '#') return 1;
  }
  ////////////////////////////////////////
  // １個下の行を見る
  ////////////////////////////////////////
  if (i < H-1) {
    // 真下を見る
    if (A[age][i+1][j] == '#') return 1;
  }
  ////////////////////////////////////////
  // カレント行を見る
  ////////////////////////////////////////
  // 左を見る
  if (j > 0) {
    if (A[age][i][j-1] == '#') return 1;
  }
  // 右を見る
  if (j < W-1) {
    if (A[age][i][j+1] == '#') return 1;
  }
  return 0;
}

int main(void)
{
  int H, W, i, j, age, change;
  scanf("%d %d", &H, &W);
//printf("H(%d) W(%d)\n", H, W);
  age = 0;
  for (i = 0; i < H; ++i) {
    scanf("%s", A[age][i]);
  }
//printf("%s\n", A[age][i]);
  //dump_A(age, H, W);
  if (is_all_black(age, H, W)) return 0;
  for (;;) {
    ++age;
    change = 0;
    memcpy(&A[age], &A[age-1], H*1001);
    //dump_A(age, H, W);
    for (i = 0; i < H; ++i) {
      for (j = 0; j < W; ++j) {
        if (A[age-1][i][j] == '#') continue;
        if (near_black(age-1, i, j, H, W) == 1) {
          A[age][i][j] = '#';
          change++;
        }
      }
    }
    if (change == 0) {
      printf("%d\n", age - 1);
      return 0;
    }
  }
}