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(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  int h,w;
  cin>>h>>w;
  vector<string> g(h);
  for(int i=0;i<h;i++)cin>>g[i];
  vector<vector<int>> d(h,vector<int>(w,-1));
  vector<pair<int,int>> q;
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      if(g[i][j]=='#'){
        q.emplace_back(i,j);
        d[i][j]=0;
      }
    }
  }
  for(int b=0;b<(int)q.size();b++){
    for(int dx=-1;dx<=1;dx++){
      for(int dy=-1;dy<=1;dy++){
        if(dx*dx+dy*dy==1){
          int xk=q[b].first+dx;
          int yk=q[b].second+dy;
          if(xk>=0 && yk>=0 && xk<h && yk<w && d[xk][yk]==-1){
            q.emplace_back(xk,yk);
            d[xk][yk]=d[q[b].first][q[b].second]+1;
          }
        }
      }
    }
  }
  cout << d[q.back().first][q.back().second] << '\n';
  return 0;
}
