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<iostream>
#include<string>
#include<queue>
using namespace std;
#define N 1005
#define pii pair<int,int>
#define mp make_pair
const int INF=1<<29;
const int dx[]={1,-1,0,0};
const int dy[]={0,0,1,-1};

int H,W,ans;
char s[N][N];
int d[N][N];
queue<pii> q;
bool inRange(int x,int y)
{
	return x>=0 && y>=0 &&x<H && y<W;
}

int main()
{
	
	scanf("%d%d",&H,&W);
	for(int i=0;i<H;i++)
		scanf("%s",&s[i]);
	for(int i=0;i<H;i++)
	for(int j=0;j<W;j++)
		if(s[i][j]=='.') d[i][j]=INF;
		else q.push(mp(i,j));
	
	while(!q.empty())
	{
		pii p=q.front();
		q.pop();
		for(int i=0;i<4;i++)
		{
			int nx=p.first+dx[i];
			int ny=p.second+dy[i];
			if(d[nx][ny]!=INF) continue;
			if(!inRange(nx,ny)) continue;
			d[nx][ny]=d[p.first][p.second]+1;
			ans=max(ans,d[nx][ny]);
			q.push(mp(nx,ny));
		}
	}
	cout<<ans<<'\n';
	
	return 0;
}