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 dx[5]={-1,1,0,0};
int dy[5]={0,0,1,-1};
int n,m;
string s[1005];
int cnt1=0;
int cnt2=0;
int ans=0;
queue<pair<int,int> > q;
bool check(int x,int y)
{
	if(x>=0 and x<n and y>=0 and y<m and s[x][y]=='.')
	{
		return true;
	}
	return false;
}
int main()
{
	cin>>n>>m;
	for(int i=0;i<n;i++)
	{
		cin>>s[i];
	}
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			if(s[i][j]=='#')
			{
				cnt1++;
				q.push(make_pair(i,j));
			}
		}
	}
	while(cnt1>=1)
	{
		cnt1--;
		int x=q.front().first;
		int y=q.front().second;
		q.pop();
		for(int i=0;i<4;i++)
		{
			int x1=x+dx[i];
			int y1=y+dy[i];
			if(check(x1,y1))
			{
				cnt2++;
				s[x1][y1]='#';
				q.push(make_pair(x1,y1));
			}
		}
		if(cnt1==0)
		{
			cnt1=cnt2;
			cnt2=0;
			ans++;
		}
	}
	cout<<ans-1;
	return 0;
}