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;
int n,m;
char grid[1200][1200];
queue<pair<pair<int,int>,int> > q; 
bool blk[1200][1200];
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
int main(){
	//freopen("input.txt","r",stdin);
	//freopen("output.txt","w",stdout);
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		scanf("%s",grid[i]+1);
		for(int j=1;j<=m;j++){
			if(grid[i][j]=='#'){
				q.push(make_pair(make_pair(i,j),0));
				blk[i][j]=1;
			}
		}
	}
	while(!q.empty()){
		int stp=q.front().second;
		pair<int,int> pos=q.front().first;
		q.pop();
		for(int i=0;i<4;i++){
			int nx=pos.first+dx[i];
			int ny=pos.second+dy[i];
			if(nx>=1&&nx<=n&&ny>=1&&ny<=m){
				if(!blk[nx][ny]){
					q.push(make_pair(make_pair(nx,ny),stp+1));
					blk[nx][ny]=1;
				}
			}
		}
		if(q.empty())
		{
			cout<<stp<<endl;
			return 0; 
		}
	}
	return 0;
}
