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>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

typedef long long LL;

const int maxn=1005;

int h,w;
char s[maxn][maxn];

int d[maxn*maxn][2],dis[maxn][maxn],fx[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int main()
{
	scanf("%d %d",&h,&w);
	fo(i,1,h) scanf("%s",s[i]+1);
	
	int zbj=0;
	fo(i,1,h)
		fo(j,1,w) if (s[i][j]=='#') d[++zbj][0]=i, d[zbj][1]=j, dis[i][j]=1;
	for(int zbi=1; zbi<=zbj; zbi++)
	{
		fo(k,0,3)
		{
			int xx=d[zbi][0]+fx[k][0], yy=d[zbi][1]+fx[k][1];
			if (xx>0 && xx<=h && yy>0 && yy<=w && !dis[xx][yy])
			{
				d[++zbj][0]=xx, d[zbj][1]=yy;
				dis[xx][yy]=dis[d[zbi][0]][d[zbi][1]]+1;
			}
		}
	}
	
	int ans=0;
	fo(i,1,h)
		fo(j,1,w) ans=max(ans,dis[i][j]-1);
	printf("%d\n",ans);
}