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;
int N_MAX = 1000;
int INF = 10000;
typedef pair<int,int> P;
int main()
{
  int h,w,i,j,ni,nj,res=0;
  string s,t;
  int d[N_MAX][N_MAX];
  int di[4]={1,0,-1,0}, dj[4]={0,1,0,-1};
  P p;
  cin >> h >> w;
  queue<P> que;
  for(i=0;i<h;i++){
    cin >> s;
    for(j=0;j<w;j++){
      t=s.substr(j,1);
      if(t==".")
        d[i][j]=INF;
      else{
        d[i][j]=0;
        que.push(P(i,j));
      }
    }
  }
  while(que.size()){
    p=que.front();
    que.pop();
    for(i=0;i<4;i++){
      ni=p.first+di[i];
      nj=p.second+dj[i];
      if(ni>=0 && ni<h && nj>=0 && nj<w && d[ni][nj]==INF){
        d[ni][nj]=res=d[p.first][p.second]+1;
        que.push(P(ni,nj));
      }
    }
  }
  cout << res << endl;
  return 0;
}
