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 <bits/stdc++.h>

using namespace std;

int main(){
   cin.tie(nullptr);
  ios::sync_with_stdio(false);

  int h,w;
  cin >> h >> w;
  vector<string> a(h);
  for(int i=0;i<h;i++) cin >> a[i];
  
  vector<vector<int>> d(h,vector<int>(w,-1));
  vector<pair<int,int>> que;
  
  for(int i=0;i<h;i++){
   for(int j=0;j<w;j++){
    if(a[i][j]=='#') que.emplace_back(i,j), d[i][j]=0; 
   }
  }
  
  
  for(int i=0;i<que.size();i++){
    for(int dx=-1;dx<=1;dx++){
     for(int dy=-1;dy<=1;dy++){
      if(dx*dx+dy*dy==1){
       int x=que[i].first+dx,y=que[i].second+dy;
        if(0<=x&&x<h&&0<=y&&y<w&&d[x][y]==-1){ 
        que.emplace_back(x,y);
        d[x][y]=d[que[i].first][que[i].second]+1;  
      }
      } 
     }
  }
  }
  cout << d[que.back().first][que.back().second] << endl;
}
