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;
char iomap[1005][1005];
int step[1005][1005];
int n,m;
queue<pair<int,int> > qu;
int dx[4]={-1,1,0,0};
int dy[4]={0,0,-1,1};
bool islegal(int x,int y)
{
	return (x>=0&&x<n&&y>=0&&y<m&&step[x][y]==-1);
}
void bfs()
{
	while(!qu.empty())
	{
		int x=qu.front().first,y=qu.front().second;
		qu.pop();
		for(int i=0;i<4;i++)
		{
			int xx=x+dx[i],yy=y+dy[i];
			if(!islegal(xx,yy))
				continue;
			qu.push(make_pair(xx,yy));
			step[xx][yy]=step[x][y]+1;
		} 
	}
}
int main()
{
	cin>>n>>m;
	memset(step,-1,sizeof(step));
	for(int i=0;i<n;i++)
		for(int j=0;j<m;j++)
		{
			cin>>iomap[i][j];
			if(iomap[i][j]=='#')
			{
				qu.push(make_pair(i,j));
				step[i][j]=0;
			}
		}
	bfs();
	int ans=-1;
	for(int i=0;i<n;i++)
		for(int j=0;j<m;j++)
			ans=max(ans,step[i][j]);
	cout<<ans<<endl;
	return 0;
}