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;

#define mp make_pair
#define fr first
#define sc second

long long n,m,ky[4]={-1,0,1,0},kx[4]={0,1,0,-1};
bitset<1069> a[1069];
queue<pair<pair<long long,long long>,long long>> q;

int main()
{
	long long i,j,it,y,x,w,yy,xx;
	string s;
	
	scanf("%lld%lld",&n,&m);
	for(i=1;i<=n;i++)
	{
		cin>>s;
		for(j=1;j<=m;j++)
		{
			if(s[j-1]=='.')
			{
				a[i][j]=0;
			}
			else
			{
				a[i][j]=1;
				q.push(mp(mp(i,j),0));
			}
		}
	}
	for(;!q.empty();)
	{
		y=q.front().fr.fr;
		x=q.front().fr.sc;
		w=q.front().sc;
		q.pop();
		for(it=0;it<4;it++)
		{
			yy=y+ky[it];
			xx=x+kx[it];
			if(yy>0&&xx>0&&yy<=n&&xx<=m&&!a[yy][xx])
			{
				q.push(mp(mp(yy,xx),w+1));
				a[yy][xx]=1;
			}
		}
	}
	printf("%lld\n",w);
}