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;
const int nmax=1000;
char mp[nmax][nmax];
int dist[nmax][nmax];
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};

int main(){
  int h,w;
  cin >> h >> w;
  for(int i=0;i<h;i++){
    cin >> mp[i];
  }
  queue<pair<int,int>> q;
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      if(mp[i][j]=='#'){
        dist[i][j]=0;
        q.push(make_pair(i,j));
      }
      else{
        dist[i][j]=-1;
      }
    }
  }
  while(!q.empty()){
    pair<int,int> p=q.front();
    q.pop();
    for(int i=0;i<4;i++){
      int x=p.first+dx[i],y=p.second+dy[i];
      if(0<=x&&x<h&&0<=y&&y<w&&dist[x][y]==-1){
        dist[x][y]=dist[p.first][p.second]+1;
        q.push(make_pair(x,y));
      }
    }
  }
  int ret=0;
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      ret=max(ret,dist[i][j]);
    }
  }
  cout << ret << endl;
}