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;
const int dx[]={1,-1,0,0};
const int dy[]={0,0,1,-1};
int n,m,ans;
char s[1111][1111];
int vis[1111][1111];
struct node
{
	int x,y,step;
}nex,now;
queue<node> que;
void bfs(){
	while(!que.empty()){
		now=que.front();que.pop();
		ans=max(ans,now.step);
		for(int i=0;i<4;i++){
			nex.x =now.x+dx[i];nex.y=now.y+dy[i];nex.step=now.step+1;
		
			if(s[nex.x][nex.y]=='.'&&!vis[nex.x][nex.y]){
				s[nex.x][nex.y]=='#';vis[nex.x][nex.y]=1;
				que.push(nex);
			}
		}
	}
}

int main(){

  scanf("%d%d",&n,&m);
  for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
    {
    	cin>>s[i][j];
    	now.x=i,now.y=j,now.step=0;
    	if(s[i][j]=='#')vis[i][j]=1,que.push(now);
    }
    bfs();
    printf("%d\n",ans);
	return 0;
}
