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 R,C,x,y,MAX=0;cin>>R>>C;
  int N[R][C];char M[R][C];
  queue<pair<int,int>>X;
  for(int i=0;i<R;i++)for(int j=0;j<C;j++){
    N[i][j]=R*C;cin>>M[i][j];
    if(M[i][j]=='#'){
      X.push(make_pair(i,j));
      N[i][j]=0;
    }
  }
  while(X.size()){
    x=X.front().first;y=X.front().second;
    if(x<R-1&&M[x+1][y]=='.'&&N[x+1][y]==R*C){X.push(make_pair(x+1,y));N[x+1][y]=N[x][y]+1;}
    if(x>0&&M[x-1][y]=='.'&&N[x-1][y]==R*C){X.push(make_pair(x-1,y));N[x-1][y]=N[x][y]+1;}
    if(y<C-1&&M[x][y+1]=='.'&&N[x][y+1]==R*C){X.push(make_pair(x,y+1));N[x][y+1]=N[x][y]+1;}
    if(y>0&&M[x][y-1]=='.'&&N[x][y-1]==R*C){X.push(make_pair(x,y-1));N[x][y-1]=N[x][y]+1;}
    X.pop();
  }
  for(int i=0;i<R;i++)for(int j=0;j<C;j++)
    MAX=max(MAX,N[i][j]);
  cout<<MAX<<endl;
}