Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

int main(){
  int h,w;
  cin>>h>>w;
  vector<string> a(h);
  for(int i=0; i<h; ++i)cin>>a[i];

  set<pair<int,int>> black;
  for(int i=0; i<h; ++i){
    for(int j=0; j<w; ++j){
      if(a[i][j]=='#')black.insert(make_pair(i,j));
    }
  }

  vector<vector<int>> b(h, vector<int>(w, 0));
  bool flg=true;

  vector<int> dx{0, 1, 0, -1}, dy{1, 0, -1, 0};

  int count=0, ans=0;
  while(black.size()!=0){
    set<pair<int,int>> black2;
    for(pair<int,int> z:black){
      int y,x;
      for(int k=0; k<4; ++k){
        y=z.first+dy[k], x=z.second+dx[k];
        if(0<=y && y<h && 0<=x && x<w && a[y][x]=='.'){
          b[y][x]=count+1;
          a[y][x]='#';
          black2.insert(make_pair(y,x));
        }
      }
    }
    ans=count;
    ++count;
    black.swap(black2);
  }

  cout<<ans<<endl;
}