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;
typedef long long ll;
#define INF 1LL<<62
#define inf 1000000007 

ll ch[1002][1002];
queue<pair<ll,ll>>a;
int main() {
	ll h,w;
	cin>>h>>w;
	for(ll i=0;i<h;i++){
		for(ll j=0;j<w;j++){
			char now;
			cin>>now;
			if(now=='.'){
				ch[i][j]=-1;
			}
			else{
				ch[i][j]=0;
				a.push(make_pair(i,j));
			}
		}
	}
	ll dx[4]={0,0,1,-1};
	ll dy[4]={1,-1,0,0};
	ll ans=0;
	while(!a.empty()){
		pair<ll,ll>now=a.front();
		a.pop();
		ll nowy=now.first;
		ll nowx=now.second;
		ll nowc=ch[nowy][nowx];
		for(ll i=0;i<4;i++){
			ll nex=nowx+dx[i];
			ll ney=nowy+dy[i];
			if(nex<0||nex>=w||ney<0||ney>=h){
				continue;
			}
			if(ch[ney][nex]!=-1){
				continue;
			}
			ch[ney][nex]=nowc+1;
			a.push(make_pair(ney,nex));
			ans=max(ans,ch[ney][nex]);
		}
	}
	cout <<ans;
	// your code goes here
	return 0;
}