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<iostream>
#include<vector>
using namespace std;
int main(){
	int h,w;
	cin>>h>>w;
	vector<int>a(h*w);
	vector<int>q;
	vector<int>next_q;
	char tmp;
	for(int i=0;i<h*w;i++){
		cin>>tmp;
		if(tmp=='#'){
			a[i]=0;
			q.push_back(i);
		}else{
			a[i]=1;
		}
	}
	int n,i,j,cnt=0;
	while(!q.empty()){
		while(!q.empty()){
			//for(int k=0;k<q.size();k++) cout<<q[k]<<" ";cout<<endl;
			n=q[q.size()-1];
			q.pop_back();
			i=n/w;
			j=n%w;
			if(i!=0&&a[n-w]==1){
				a[n-w]=0;
				next_q.push_back(n-w);
			}
			if(i!=h-1&&a[n+w]==1){
				a[n+w]=0;
				next_q.push_back(n+w);
			}
			if(j!=0&&a[n-1]==1){
				a[n-1]=0;
				next_q.push_back(n-1);
			}
			if(j!=w-1&&a[n+1]==1){
				a[n+1]=0;
				next_q.push_back(n+1);
			}
		}
	q=next_q;
	next_q.clear();
	cnt++;
	}
	cout<<cnt-1<<endl;
	return 0;
}