Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define pii pair<int,int>
#define fi first
#define sc second
using namespace std;
const int N=1050;
const int dx[]={0,0,-1,1};
const int dy[]={-1,1,0,0};

int n,m;
char mp[N][N];
int dis[N][N],ans=0;

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		scanf("%s",mp[i]+1);
	queue<pii>q;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(mp[i][j]=='#')dis[i][j]=1,q.push(pii(i,j));
	int x,y,nx,ny;
	while(!q.empty())
	{
		x=q.front().fi,y=q.front().sc,q.pop();
		ans=max(ans,dis[x][y]);
		for(int i=0;i<4;i++)
		{
			nx=x+dx[i],ny=y+dy[i];
			if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&!dis[nx][ny])
				dis[nx][ny]=dis[x][y]+1,q.push(pii(nx,ny));
		}
	}
	printf("%d\n",ans-1);
}