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>
#define rep(i,n)for(int i=0;i<(n);i++)
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
#define EPS (1e-10)
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

char s[2000][2000];
int dx[]{1,-1,0,0},dy[]{0,0,1,-1};
int d[2000][2000];

int main(){
	int h,w;cin>>h>>w;
	queue<P>que;
	memset(d,-1,sizeof(d));
	rep(i,h){
		scanf("%s",s[i]);
		rep(j,w){
			if(s[i][j]=='#'){
				que.push(P(i,j));
				d[i][j]=0;
			}
		}
	}
	while(!que.empty()){
		auto p=que.front();que.pop();
		rep(k,4){
			int nx=p.first+dx[k],ny=p.second+dy[k];
			if(0<=nx&&nx<h&&0<=ny&&ny<w&&d[nx][ny]==-1&&s[nx][ny]=='.'){
				d[nx][ny]=d[p.first][p.second]+1;
				que.push(P(nx,ny));
			}
		}
	}
	int Max=0;
	rep(i,h)rep(j,w)Max=max(Max,d[i][j]);
	cout<<Max<<endl;
}