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 N 1000
#define M 1000
#define mp make_pair
#define X first
#define Y second
#define pb push_back
char s[N+1][M+1];
vector<pair<int,int> > blk,wb_blk;
int main(){
	const int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
	int n,m,ans=0,i,j;scanf("%d%d",&n,&m);
	for(i=1;i<=n;i++)scanf("%s",s[i]+1);
	for(i=1;i<=n;i++)for(j=1;j<=m;j++)
		if(s[i][j]=='#')blk.pb(mp(i,j));
	while(blk.size()){
		for(i=0;i<blk.size();i++){
			int x=blk[i].X,y=blk[i].Y;
			for(j=0;j<4;j++){
				int xx=x+dx[j],yy=y+dy[j];
				if(s[xx][yy]=='.')wb_blk.pb(mp(xx,yy)),s[xx][yy]='#';
			}
		}
		blk.swap(wb_blk);wb_blk.clear();
		ans++;
	}
	printf("%d\n",ans-1);
	return 0;
}
/*1
3 3
...
.#.
...
*/
/*2
6 6
..#..#
......
#..#..
......
.#....
....#.
*/