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>
using namespace std;
const int dx[]={-1,0,0,1},dy[]={0,-1,1,0};
int n,m,w[1101][1101];
char s[1101][1101];
int main()
{
	cin>>n>>m;
	for(int i=1;i<=n;i++)cin>>(s[i]+1);
	queue<pair<int,int>> q;
	memset(w,0x3f,sizeof w);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(s[i][j]=='#')w[i][j]=0,q.push({i,j});
	while(!q.empty())
	{
		int x=q.front().first,y=q.front().second;q.pop();
		for(int d=0;d<4;d++)
		{
			int nx=x+dx[d],ny=y+dy[d];
			if(nx>0&&ny>0&&nx<=n&&ny<=m&&w[nx][ny]>w[x][y]+1)w[nx][ny]=w[x][y]+1,q.push({nx,ny});
		}
	}
	int ans=0;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)ans=max(ans,w[i][j]);
	cout<<ans<<'\n';
}
