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<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,i,j,k,qx[1005*1005],qy[1005*1005],bg,ed,dis[1005][1005],f[10][5]={{-1,0},{1,0},{0,-1},{0,1}};
char s[1005][1005];
int main()
{
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;++i)
	{	
		scanf("%s",s[i]+1);
		for(j=1;j<=m;++j)
			if(s[i][j]=='#')
			{
				++ed;
				qx[ed]=i,qy[ed]=j;
			}
			else
				dis[i][j]=-1;
	}
	bg=1;
	while(bg<=ed)
	{
		for(i=0;i<4;++i)
		{
			j=qx[bg]+f[i][0],k=qy[bg]+f[i][1];
			if(j>=1&&j<=n&&k>=1&&k<=m&&dis[j][k]==-1)
			{
				dis[j][k]=dis[qx[bg]][qy[bg]]+1;
				++ed;
				qx[ed]=j,qy[ed]=k;
			}
		}
		++bg;
	}
	k=0;
	for(i=1;i<=n;++i)
		for(j=1;j<=m;++j)
			k=max(k,dis[i][j]);
	printf("%d",k);
	return 0;
}
