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>
#include<queue>
using namespace std;
int main(){
  int h,w;
  cin>>h>>w;
  queue<pair<int,int>> q;
  vector<vector<int>> d(h+2,vector<int>(w+2,0));
  for(int i=1;i<=h;i++){
    for(int j=1;j<=w;j++){
      char block;
      cin>>block;
      if(block=='.'){
        d.at(i).at(j)=1;
      }
      else q.push({i,j});
    }
  }
  int count=0;
  vector<int> dx={1,0,-1,0};
  vector<int> dy={0,1,0,-1};
  while(!q.empty()){
    queue<pair<int,int>> qq;
    while(!q.empty()){
      pair<int,int> p=q.front();
      q.pop();
        for(int l=0;l<4;l++){
          if(d.at(p.first+dx.at(l)).at(p.second+dy.at(l))){
            d.at(p.first+dx.at(l)).at(p.second+dy.at(l))=0;
            qq.push({p.first+dx.at(l),p.second+dy.at(l)});
          }
        }
    }
    count++;
    q=qq;
  }
  cout<<count-1<<endl;
}