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;
const int Maxn=1005;
int dis[Maxn][Maxn],n,m,ans;
string str;
const int dx[]={0,0,-1,0,1},dy[]={0,-1,0,1,0};
struct point
{
	int x,y,cnt;
};
queue <point> Qu;
void bfs(void)
{
	while(!Qu.empty())
	{
		point now=Qu.front();
		Qu.pop();
		now.cnt++;
		for(int i=1;i<=4;i++)
		{
			now.x+=dx[i],now.y+=dy[i];
			if(now.x==0||now.x==n+1||now.y==0||now.y==m+1) goto A;
			if(dis[now.x][now.y]==-1)
			{
				Qu.push(now);
				dis[now.x][now.y]=now.cnt;
				ans=max(ans,now.cnt);
			}
		A:	now.x-=dx[i],now.y-=dy[i];
		}
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	memset(dis,-1,sizeof(dis));
	for(int i=1;i<=n;i++)
	{
		cin>>str;
		for(int j=1;j<=m;j++)
			if(str[j-1]=='#') Qu.push((point){i,j,0}),dis[i][j]=0;
	}
	bfs();
	printf("%d",ans);
	return 0;
}