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: Time Limit Exceeded)

#include <stdio.h>

int main() {
  int H,W,i,j,flag=1,count=0;
  scanf("%d%d", &H, &W);
  char A[H][W],newA[H][W];
  int visited[H][W];
  for(i = 0; i < H; i++){
    scanf("%s",A[i]);
    for(j = 0; j < W; j++){
      visited[i][j] = 0;
      newA[i][j] = A[i][j];
      if(A[i][j] == '.'){
	flag = 0;
      }
    }
  }
  while(flag == 0){
    flag = 1;
    for(i = 0; i < H; i++){
      for(j = 0; j < W; j++){
	
	if(visited[i][j] == 0 && A[i][j] == '#'){
	  visited[i][j] = 1;
	  if(A[i-1][j] == '.' && i-1 >= 0){
	    newA[i-1][j] = '#';
	  }
	  if(A[i+1][j] == '.' && i+1 <= H-1){
	    newA[i+1][j] = '#';
	  }
	  if(A[i][j-1] == '.' && j-1 >= 0){
	    newA[i][j-1] = '#';
	  }
	  if(A[i][j+1] == '.' && j+1 <= W-1){
	    newA[i][j+1] = '#';
	  }
	}
      }
    }
    for(i = 0; i < H; i++){
      for(j = 0; j < W; j++){
	if(visited[i][j] == 0){
	  A[i][j] = newA[i][j];
	  if(newA[i][j] == '.'){
	    flag = 0;
	  }
	}
      }
    }
    count++;
  }
  printf("%d\n",count);
  return 0;
}