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 N=1007,dx[4]={0,0,1,-1},dy[4]={1,-1,0,0};
int n,m,ans,qs,qe,d[N][N],qx[N*N],qy[N*N];
char mp[N][N];
int main()
{
	cin>>n>>m;
	for(int i=1;i<=n;i++)scanf("%s",mp[i]+1);
	for(int i=1;i<=n;i++)
	for(int j=1;j<=m;j++)
	if(mp[i][j]=='.')d[i][j]=1e9;
	else d[i][j]=0,qx[qe]=i,qy[qe++]=j;
	while(qs<qe)
	{
		int x=qx[qs],y=qy[qs++];
		for(int i=0;i<4;i++)
		{
			int u=x+dx[i],v=y+dy[i];
			if(u<1||u>n||v<1||v>m||d[u][v]<=d[x][y]+1)continue;
			d[u][v]=d[x][y]+1,ans=max(ans,d[u][v]);
			qx[qe]=u,qy[qe++]=v;
		}
	}
	cout<<ans<<endl;
}