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
#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
typedef long long ll;
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
using namespace std;
const ll MAX_H=1000,MAX_W=1000;
ll H,W;
typedef pair<ll,ll> P;
char A[MAX_H][MAX_W+1];
ll dst[MAX_H][MAX_W];
const ll dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>H>>W;
	REP(i,H)cin>>A[i];
	queue<P> que;
	REP(i,H)REP(j,W){
		if(A[i][j]=='#'){
			dst[i][j]=0;
			que.push(P(i,j));
		}else{
			dst[i][j]=-1;
		}
	}
	while(!que.empty()){
		P p=que.front();
		que.pop();
		REP(i,4){
			ll i2=p.first+dy[i],j2=p.second+dx[i];
			if(i2>=0&&i2<H&&j2>=0&&j2<W&&dst[i2][j2]==-1){
				dst[i2][j2]=dst[p.first][p.second]+1;
				que.push(P(i2,j2));
			}
		}
	}
	ll ans=0;
	REP(i,H)REP(j,W)ans=max(ans,dst[i][j]);
	cout<<ans<<endl;
}