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;
  int ans=0;
  cin>>h>>w;
  char a[h][w];
  queue<pair<int,int>> q;
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      cin>>a[i][j];
      if(a[i][j]=='#'){q.push(pair<int,int> (i,j));}
    }
  }
  vector<vector<int>> distance(h, vector<int> (w, 0));
  while(q.size()){
    pair<int,int> start=q.front(); q.pop();
    int x=start.first;
    int y=start.second;
    int dx[]={1,0,-1,0};
    int dy[]={0,1,0,-1};
    for(int i=0;i<4;i++){
      int cx=x+dx[i];
      int cy=y+dy[i];
      if(0<=cx&&cx<h&&0<=cy&&cy<w&&a[cx][cy]=='.'&&distance[cx][cy]==0){
        distance[cx][cy]=distance[x][y]+1;
        q.push(pair<int,int>(cx,cy));
      }
    }
  }
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      ans=max(ans,distance[i][j]);
    }
  }
  cout<<ans<<endl;
}