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<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int dx[4]={0,0,1,-1};
const int dy[4]={1,-1,0,0};
int n,m,dis[1005][1005];
char s[1005][1005];
queue<int>qx,qy;
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
	{
		scanf("%s",s[i]+1);
		for(int j=1;j<=m;j++)
		{
			if(s[i][j]=='#')
			{
				dis[i][j]=1;
				qx.push(i);
				qy.push(j);
			}
		}
	}
	int ans=1;
	while(!qx.empty())
	{
		int x=qx.front();
		int y=qy.front();
		qx.pop(),qy.pop();
		for(int k=0;k<4;k++)
		{
			int xx=x+dx[k],yy=y+dy[k];
			if(s[xx][yy]=='.')
			{
				s[xx][yy]='#';
				dis[xx][yy]=dis[x][y]+1;
				ans=max(ans,dis[xx][yy]);
				qx.push(xx),qy.push(yy);
			}
		}
	}
	printf("%d\n",ans-1);
	return 0;
}