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: Runtime Error)

#include <stdio.h>

int main(void){
    int H,W;
     	scanf("%d", &H);//数値の取得
  		scanf("%d", &W);
  char a[H][W];
  int i,j;
   for(i=0;i<H ;i++){
        scanf("%s", a[i]);//数値の取得
   }
  int count = 0;
  int check = 0;
  int change[H][W];
  
  do{
    check = 0;
  for(i=0;i<H;i++){
    for(j=0;j<W;j++){
    if(a[i][j] == '#'){
      change[i][j] = 1;
    }
      if(a[i][j] == '.'){
        check = 1;
      }
    }
  }
    for(i=0;i<H;i++){
    for(j=0;j<W;j++){
    if(change[i][j] == 1){
      if(i!=0){
    	a[i-1][j] = '#';
       // printf("test:%c\n",a[i-1][j]);
      }
      if(i != W-1){
        //printf("suuc2\n");
     	a[i+1][j] = '#';
        //printf("testi+1,j:%c\n",a[i+1][j]);
      }
      if(j != 0){
       // printf("suuc3\n");        
     	a[i][j-1] = '#';
       // printf("test2:%c\n",a[i][j-1]);
      }
      if(j != H-1){
     	//printf("suuc4\n");        
        a[i][j+1] = '#';
       // printf("testi,j+1:%c",a[i][j+1]);
      }
    }
    }
    }
     count++;
  }while(check == 1);
 
    printf("%d",count-1);
    return 0;
}