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=1e3+1,X[4]={-1,0,0,1},Y[4]={0,-1,1,0};
bool v[N][N];
int n,m,p[N*N],q[N*N],s,t,P,Q,f[N][N],A;
string S[N];
int main()
{
	ios::sync_with_stdio(false);
	cin>>n>>m;
	for(int i=1;i<=n;i++)
		cin>>S[i];
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(S[i][j-1]=='#')
				p[++t]=i,q[t]=j,v[i][j]=true;
	while(s<t)
	{
		P=p[++s],Q=q[s];
		for(int i=0;i<4;i++)
			if(1<=P+X[i]&&P+X[i]<=n&&1<=Q+Y[i]&&Q+Y[i]<=m&&!v[P+X[i]][Q+Y[i]])
				p[++t]=P+X[i],q[t]=Q+Y[i],f[P+X[i]][Q+Y[i]]=f[P][Q]+1,v[P+X[i]][Q+Y[i]]=true;
	}
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			A=max(A,f[i][j]);
	cout<<A;
	return 0;
}