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 <cmath>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <iostream>
#include <bitset>
using namespace std;
#define N 1005
int dis[N][N],ans,n,m;char a[N][N];
int dx[]={1,0,-1,0},dy[]={0,1,0,-1};
void bfs()
{
	memset(dis,-1,sizeof(dis));
	queue<pair<int ,int > >q;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(a[i][j]=='#')dis[i][j]=0,q.push(make_pair(i,j));
	while(!q.empty())
	{
		int x=q.front().first,y=q.front().second;q.pop();ans=max(ans,dis[x][y]);
		for(int i=0;i<4;i++)
		{
			int tx=x+dx[i],ty=y+dy[i];
			if(tx<=n&&tx>0&&ty<=m&&ty>0&&dis[tx][ty]==-1)
				dis[tx][ty]=dis[x][y]+1,q.push(make_pair(tx,ty));
		}
	}
	printf("%d\n",ans);
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)scanf("%s",a[i]+1);bfs();
}