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<iostream>
#include<queue>
using namespace std;
int H,W;
string s[1000];
int dis[1000][1000];
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
int main()
{
	cin>>H>>W;
	for(int i=0;i<H;i++)cin>>s[i];
	queue<pair<int,int> >P;
	for(int i=0;i<H;i++)
	{
		for(int j=0;j<W;j++)
		{
			if(s[i][j]=='#')
			{
				dis[i][j]=1;
				P.push(make_pair(i,j));
			}
		}
	}
	while(!P.empty())
	{
		int x=P.front().first,y=P.front().second;
		P.pop();
		for(int r=0;r<4;r++)
		{
			int tx=x+dx[r],ty=y+dy[r];
			if(tx<0||ty<0||tx>=H||ty>=W||dis[tx][ty]>0)continue;
			dis[tx][ty]=dis[x][y]+1;
			P.push(make_pair(tx,ty));
		}
	}
	int ans=0;
	for(int i=0;i<H;i++)for(int j=0;j<W;j++)
	{
		if(ans<dis[i][j])ans=dis[i][j];
	}
	cout<<ans-1<<endl;
}