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 <algorithm>
#include <string>
#include <queue>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;


int main(void){
  int h,w,i,j;
  string g[1005];
  queue<P> q;
  
  cin >> h >> w;
  for (i=0;i<h;i++) cin >> g[i];
  
  for (i=0;i<h;i++) for (j=0;j<w;j++){
    if (g[i][j]=='#') q.push(P(i,j));
  }
  
  int qs,t=0,ni,nj;
  int di[4]={1,0,-1,0},dj[4]={0,1,0,-1};
  P p;
  while (q.size()>0){
    qs=q.size();
    
    for (i=0;i<qs;i++){
      p=q.front();
      q.pop();
      for (j=0;j<4;j++){
        ni=p.first+di[j];
        nj=p.second+dj[j];
        if (ni>=0 && ni<h && nj>=0 && nj<w && g[ni][nj]=='.'){
          g[ni][nj]='#';
          q.push(P(ni,nj));
        }
      }
    }
    
    t++;
    
  }
  
  cout << t-1 << endl;
  
  return 0;
}