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;
using ll = long long;
const ll mod = 998244353;
vector<ll>r{-1,0,1,0};
vector<ll>l{0,-1,0,1};

int main(){
  ll h,w;
  cin >> h >> w;
  vector<vector<ll>>length(h,vector<ll>(w,-1));
  queue<pair<ll,ll>>que;
  for(int i=0; i<h; i++){
   for(int j=0; j<w; j++){
     char a;
     cin >> a;
     if(a=='#'){
       length[i][j]=0;
       que.push({i,j});
     }
   }
  }
  ll MAX=0;
  while(que.size()>0){
    ll x=que.front().first,y=que.front().second;
    que.pop();
    for(int i=0; i<4; i++){
      if(x+r[i]<0 || x+r[i]>=h || y+l[i]<0 || y+r[i]>=w) continue;
      if(length[x+r[i]][y+l[i]]!=-1) continue;
      length[x+r[i]][y+l[i]]=length[x][y]+1;
      que.push({x+r[i],y+l[i]});
      MAX=max(MAX,length[x+r[i]][y+l[i]]);
    }
  }
  
  cout << MAX << endl;
  
}
  
  
  
  
