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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<vector>
#include<stdlib.h>
#include<time.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
#define MAX 1000000000
int dis[1050][1050];

/*黒マスの位置*/
typedef struct position {
 int y;
 int x;
}position_ ;

int H,W;
int dy[4]={0, 1,-1,0};
int dx[4]={-1,0,0,1};
int ans=0;

int main(void){
 string A[1050];
 queue<position_> Q;
 int cnt=0;
 cin>>H>>W;
 
 for(int i=0; i<H; i++)cin>>A[i];
 
 position_ B;
 for(int i=0; i<H; i++)
  for(int j=0; j<W; j++){
   if(A[i][j]=='#'){
    B.x=j;
    B.y=i;
    Q.push(B);
    dis[i][j]=0;
   }
   else{
    dis[i][j]=MAX;
   }
  }
 
 position_ C;
 while(Q.empty()!=true){
  B=Q.front();
  Q.pop();
  for(int i=0; i<4; i++){
   if(0<=dx[i]+B.x && dx[i]+B.x<W && 0<=dy[i]+B.y && dy[i]+B.y<H){
    if(dis[dy[i]+B.y][dx[i]+B.x]==MAX){
     dis[dy[i]+B.y][dx[i]+B.x]=dis[B.y][B.x]+1;
     C.x=B.x+dx[i];
     C.y=B.y+dy[i];
     Q.push(C);
     ans=max(ans,dis[dy[i]+B.y][dx[i]+B.x]);
    }
   }
  }
 }
 cout<<ans<<endl;
 return 0;
}