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;
int dx[]={1,-1,0,0},dy[]={0,0,1,-1};
int h,w;
int main()
{
	cin>>h>>w;
	string s[1005];
	for(int i=0;i<h;i++)
		cin>>s[i];
	int d[1005][1005];
	queue <pair <int,int> > que ;
	for(int i=0;i<1005;i++)
		memset(d[i],-1,sizeof(d[i]));
	for(int i=0;i<h;i++)
	{
		for(int j=0;j<w;j++)
		{
			if(s[i][j]=='#')
			{
				d[i][j]=0;
				que.push(make_pair(i,j));
			}
		}
	}
	while(!que.empty())
	{
		pair <int,int> p=que.front();
		que.pop();
		int x=p.first,y=p.second;
		for(int k=0;k<4;k++)
		{
			int vx=x+dx[k],vy=y+dy[k];
			if(vx<0 || vx>=h || vy<0 || vy>=w || d[vx][vy]!=-1)
				continue;
			d[vx][vy]=d[x][y]+1;
			que.push(make_pair(vx,vy));
		}
	}
	int ans=0;
	for(int i=0;i<h;i++)
	{
		for(int j=0;j<w;j++)
		{
			ans=max(ans,d[i][j]);
		}
	}
	cout<<ans<<"\n";
 	return 0;
}
