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>
#define MOD 1000000007
#define F first
#define S second
using namespace std;
typedef pair<int,int> pii;
int n,k,arr[101010];

int main()
{
	cin>>n>>k;
	for(int i=0;i<n;i++)
	{
		scanf("%d",arr+i);
	} sort(arr,arr+n);
	
}*/
#include <bits/stdc++.h>
#define MOD 1000000007
#define F first
#define S second
using namespace std;
typedef pair<int,int> pii;
int n,k,arr[101010];
double enclosing_sphere(vector<double> x, vector<double> y){
  int n = x.size();
  auto hyp = [](double x, double y){
    return x * x + y * y;
  };
  double px = 0, py = 0;
  for(int i=0; i<n; i++){
    px += x[i];
    py += y[i];
  }
  px *= 1.0 / n;
  py *= 1.0 / n;
  double rat = 0.1, maxv;
  for(int i=0; i<10000; i++){
    maxv = -1;
    int maxp = -1;
    for(int j=0; j<n; j++){
      double tmp = hyp(x[j] - px, y[j] - py);
      if(maxv < tmp){
        maxv = tmp;
        maxp = j;
      }
    }
    px += (x[maxp] - px) * rat;
    py += (y[maxp] - py) * rat;
    rat *= 0.998;
  }
  return sqrt(maxv);
}
int main()
{
	cin>>n; vector<double> xx,yy;
	for(int i=0;i<n;i++)
	{
		double p,q;
		scanf("%lf %lf",&p,&q);
		xx.push_back(p); yy.push_back(q);
	} printf("%lf",enclosing_sphere(xx,yy));
	
}