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>
#define pii pair<int,int>
using namespace std;
const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};
int n,m,cnt;
queue<pii>q;
char c[1010][1010];
int main()
{
	cin>>n>>m;
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			cin>>c[i][j];
			if(c[i][j]=='#')
			{
				q.push(pii(i,j));
			}
		}
	}
	while(!q.empty())
	{
		queue<pii>qq=q;
		while(!q.empty())
		{
			q.pop();
		}
		cnt++;
		while(!qq.empty())
		{
			pii p=qq.front();
			qq.pop();
			int x=p.first,y=p.second;
			for(int i=0;i<4;i++)
			{
				int xx=x+dx[i],yy=y+dy[i];
				if(xx>=0 && xx<n && yy>=0 && yy<m && c[xx][yy]=='.')
				{
					q.push(pii(xx,yy));
					c[xx][yy]='#';
				}
			}
		}
	}
	cout<<cnt-1<<endl;
	return 0;
}