Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define fi first
#define se second

using namespace std;
char arr[1005][1005];
int r,c,ans,dis[1005][1005];
int x[]={1,-1,0,0};
int y[]={0,0,-1,1};
pair<int,int> now;
queue <pair<int,int> > q;
void bfs(){
	while(!q.empty()){
		now=q.front();
		q.pop();
		for(int i =0;i<=3;i++){
			int nextx=now.fi+x[i];
			int nexty=now.se+y[i];
			if(dis[nextx][nexty]>dis[now.fi][now.se]+1){
				dis[nextx][nexty]=dis[now.fi][now.se]+1;
				q.push({nextx,nexty});
			}
		}
	}
}
int main(){
	cin>>r>>c;
	for(int i =1;i<=r;i++){
		for(int j=1;j<=c;j++){
			dis[i][j]=1e9;
		}
	}
	for(int i =1;i<=r;i++){
		for(int j =1;j<=c;j++){
			cin>>arr[i][j];
			if(arr[i][j]=='#'){
				dis[i][j]=0;
				q.push({i,j});
			}
		}
	}

	bfs();
	for(int i =1;i<=r;i++){
		for(int j=1;j<=c;j++){
			ans=max(ans,dis[i][j]);
		}
	}
	cout<<ans<<endl;
}