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.

Examples

Input

3 3
...
.#.
...


Output

2


Input

6 6
..#..#
......
..#..
......
.#....
....#.


Output

3
#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
typedef pair<int,int>P;
typedef pair<int,P>PP;
int main(){
  int h,w;
  int dy[4]={-1,0,1,0};
  int dx[4]={0,1,0,-1};
  char maps[1005][1005];
  queue<PP>que;
  
  cin>>h>>w;
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      cin>>maps[i][j];
      if(maps[i][j]=='#'){
        que.push(PP(0,P(i,j)));
      }
    }
  }
  
  int maxim=0;
  while(!que.empty()){
    PP now=que.front();que.pop();
    int num=now.first;
    int y=now.second.first,x=now.second.second;
    for(int i=0;i<4;i++){
      int ny=y+dy[i];
      int nx=x+dx[i];
      if(maps[ny][nx]=='.'&&ny>=0&&ny<=h-1&&nx>=0&&nx<=w-1){
        maxim=max(maxim,num+1);
        que.push(PP(num+1,P(ny,nx)));
        maps[ny][nx]='#';
      }
    }
  }
  
  cout<<maxim<<endl;
  return(0);
}
