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;
#define rep(i,n) for(int i=0;i<(n);++i)
int h,w,x,y,nx,ny,t,r,a[1010][1010],dx[]={0,1,0,-1},dy[]={1,0,-1,0};
char c;
int main(){
  queue<tuple<int,int,int>>que;
  cin>>h>>w;
  rep(i,h)rep(j,w){
    cin>>c;
    a[i][j]=99999999;
    if(c=='#')que.emplace(i,j,0);
  }
  while(!que.empty()){
    tie(y,x,t)=que.front();que.pop();
    if(a[y][x]>t){
      a[y][x]=t;
      r=max(r,t);
      rep(i,4){
        ny=y+dy[i];nx=x+dx[i];
        if(nx<0||ny<0||nx>=w||ny>=h)continue;
        que.emplace(ny,nx,t+1);
      }
    }
  }
  cout<<r;
}