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>
using namespace std;
int n,m,ans;
struct Node{
	int x,y;
};
queue<Node> q;
int p[1010][1010];
void bfs()
{
	while(!q.empty())
	{
		Node a=q.front();
		q.pop();
		for (int dx = -1; dx <= 1; dx++)
			for (int dy = -1; dy <= 1; dy++)
				if (dx * dx + dy * dy == 1)
				{
					Node b;
					b.x=a.x+dx;
					b.y=a.y+dy;
					if(b.x>0&&b.y>0&&b.x<=n&&b.y<=m&&p[b.x][b.y]==-1)
					{
						q.push(b);
						p[b.x][b.y]=p[a.x][a.y]+1;
						//cout<<b.x<<" "<<b.y<<" "<<p[b.x][b.y]<<endl;
					}
				}
		ans=p[a.x][a.y];
	}
}
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	cin>>n>>m;
	string s;
	for(int i=1;i<=n;i++)
	{
		cin>>s;
		for(int j=1;j<=m;j++)
		{
			if(s[j-1]=='#')
			{
				Node xx;
				xx.x=i,xx.y=j;
				q.push(xx);
			}else{
				p[i][j]=-1;
			}
		}
	}
	bfs();
	cout<<ans<<endl;
	return 0;
}