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;

char pic[1010][1010];
int mk[1010][1010];
int a[]={-1,1,0,0},b[]={0,0,-1,1};
queue<pair<int,pair<int,int> > > q;

int main()
{
	int n,m;
	scanf("%d %d",&n,&m);
	for(int i=0;i<n;i++)
	{
		scanf("%s",pic[i]);
		for(int j=0;j<m;j++)
		{
			if(pic[i][j]=='#')
			{
				q.push(make_pair(0,make_pair(i,j)));
			}
		}
	}
	while(!q.empty())
	{
		//printf("d\n");
		int p1=q.front().second.first,p2=q.front().second.second;
		int val=q.front().first;
		q.pop();
		for(int i=0;i<4;i++)
		{
			int g1=p1+a[i],g2=p2+b[i];
			if(g1<0||g1>=n||g2<0||g2>=m)
			{
				continue;
			}
			if(pic[g1][g2]=='.')
			{
				pic[g1][g2]='#';
				mk[g1][g2]=val+1;
				q.push(make_pair(val+1,make_pair(g1,g2)));
			}
		}
	}
	int ans=0;
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			ans=max(ans,mk[i][j]);
		}
	}
	printf("%d",ans);
}