Given are N points (x_i, y_i) in a two-dimensional plane.

Find the minimum radius of a circle such that all the points are inside or on it.

Constraints

* 2 \leq N \leq 50
* 0 \leq x_i \leq 1000
* 0 \leq y_i \leq 1000
* The given N points are all different.
* The values in input are all integers.

Input

Input is given from Standard Input in the following format:


N
x_1 y_1
:
x_N y_N


Output

Print the minimum radius of a circle such that all the N points are inside or on it.

Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.

Examples

Input

2
0 0
1 0


Output

0.500000000000000000


Input

3
0 0
0 1
1 0


Output

0.707106781186497524


Input

10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9


Output

6.726812023536805158
#include <bits/stdc++.h>
using namespace std;
#define pi 3.14159265359
#define int long long
#define pii pair<int,int>
const int mod=1e9 + 7;
 
int dx[]={-1, 0, 1, 0};
int dy[]={0, -1, 0, 1};
 

double dist(double a, double b) {
	return a*a + b*b;
}

int32_t main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);

   int n;  cin>>n;
  vector<double>x(1005), y(1005) ;
   double X,Y,d,e;  
  
   cout<<setprecision(15)<<fixed;
  for (int i=0;i<n;i++){
		cin>>x[i]>>y[i];
		X += x[i]; Y += y[i];
	}
	X/=n; Y/=n;
	double P=0.1;
	for(int i=0;i<30000;i++) {
		int f=0;
		d =dist(X-x[0],Y-y[0]);
		for(int j=1;j<n;j++){
			e = dist(X-x[j],Y-y[j]);
			if(d<e)
            {d=e; f=j;}
		}
		X+=(x[f]-X)*P;
		Y+=(y[f]-Y)*P;
		P*=0.999;
	}
	cout<<sqrt(d);

}


