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 dist[1010][1010],H,W;
queue<int> q;
const int dx[]={0,0,1,-1},dy[]={-1,1,0,0};
int getfront()
{
	int res=q.front();
	q.pop();
	return res;
}
int main()
{
	cin>>H>>W;
	for(int i=0;i<H;i++)
	{
		string s;
		cin>>s;
		for(int j=0;j<W;j++)
			if(s[j]=='#')
			{
				dist[i][j]=1;
				q.push(i);
				q.push(j);
			}
	}
	while(!q.empty())
	{
		int x=getfront(),y=getfront();
		for(int i=0;i<4;i++)
		{
			int tox=x+dx[i],toy=y+dy[i];
			if(tox<0||tox>=H||toy<0||toy>=W||dist[tox][toy])
				continue;
			dist[tox][toy]=dist[x][y]+1;
			q.push(tox);
			q.push(toy);
		}
	}
	int B=-1;
	for(int i=0;i<H;i++)
		for(int j=0;j<W;j++)
			B=max(B,dist[i][j]);
	cout<<B-1<<endl;
	return 0;
}