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;
typedef pair<int,int> P;
int main(void){
  int h,w,i,j,a,b,m=1;
  char c;
  vector<int> di{1,0,-1,0},dj{0,1,0,-1};
  cin >> h >> w;
  vector<vector<int>> f(h+3,vector<int>(w+3,0));
  queue<P> q;
  for (i=1;i<=h;i++) for (j=1;j<=w;j++){
    cin >> c;
    if (c=='#'){
      f.at(i).at(j)=1;
      q.push(P(i,j));
    }
  }
  while (q.size()){
    a=q.front().first;
    b=q.front().second;
    q.pop();
    if (m<f.at(a).at(b)) m=f.at(a).at(b);
    for (i=0;i<4;i++) if (a+di.at(i)>0 && a+di.at(i)<=h && b+dj.at(i)>0 && b+dj.at(i)<=w){
      if (f.at( a + di.at(i) ).at( b + dj.at(i) ) == 0){
        f.at( a + di.at(i) ).at( b + dj.at(i) ) = f.at(a).at(b) + 1;
        q.push( P( a + di.at(i) , b + dj.at(i) ) );
      }
    }
  }
  cout << m-1 << endl;
  return 0;
}