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;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define P pair<int,int>
int h,w;
char c[1111][1111];
queue<P>q;
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};
int d[1111][1111];
signed main(){
	cin>>h>>w;
	rep(i,h)rep(j,w)d[i][j]=1145141919;
	rep(i,h){
		rep(j,w){
			cin>>c[i][j];
			if(c[i][j]=='#'){
				q.push({i,j});
				d[i][j]=0;
			}
		}
	}
	
	while(!q.empty()){
		P p=q.front();
		q.pop();
		rep(i,4){
			int X=p.first+dx[i];
			int Y=p.second+dy[i];
			if(0<=X&&X<h&&0<=Y&&Y<w){
				if(d[X][Y]>d[p.first][p.second]+1){
					d[X][Y]=d[p.first][p.second]+1;
					q.push({X,Y});
				}
			}
		}
	}
	
	int ans=0;
	rep(i,h)rep(j,w)ans=max(ans,d[i][j]);
	cout<<ans<<endl;
	return 0;
}
