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;

vector<pair<int,int> > v,V;
int a[1011][1011],d[4][2],n,m;
char s[1010];

int main()
{
	scanf("%d%d",&n,&m);
	memset(a,0,sizeof(a));
	for (int i=1; i<=n; i++)
	{
		scanf("%s",s);
		for (int j=0; j<m; j++) a[i][j+1]=(s[j]=='.');
	}
	d[0][0]=d[1][0]=d[2][1]=d[3][1]=0;
	d[0][1]=d[2][0]=1,d[1][1]=d[3][0]=-1;
	for (int i=1; i<=n; i++)
		for (int j=1; j<=m; j++)
			if (!a[i][j]) v.push_back(make_pair(i,j));
	for (int ans=-1; ; ans++)
	{
		int siz=v.size();
		if (!siz) return printf("%d\n",ans),0;
		V.clear();
		for (int i=0; i<siz; i++)
		{
			int x=v[i].first,y=v[i].second;
			for (int j=0; j<4; j++)
				if (a[x+d[j][0]][y+d[j][1]]) a[x+d[j][0]][y+d[j][1]]=0,V.push_back(make_pair(x+d[j][0],y+d[j][1]));
		}
		v=V;
	}
	return 0;
}