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;

const double EPS = 1e-8;
const complex<double> I = complex<double>(0.0, 1.0);

int main(){
  int N;
  cin >> N;
  
  vector<complex<double>> p(N);
  for(int i = 0; i < N; i++){
    double re, im;
    cin >> re >> im;
    p[i] = complex<double>(re, im);
  }
  
  double t = 2000.0;
  double b = 0.0;
  double R = -1.0;
  for(int k = 0; k < 50; k++){
    R = (t + b) / 2.0;
    vector<complex<double>> intersections;
    for(int i = 0; i < N - 1; i++){
      for(int j = i + 1; j < N; j++){
        double d = abs(p[i] - p[j]);
        if(d > 2.0 * R + EPS) continue;
        complex<double> dir = (p[i] - p[j]) / d * sqrt(R * R + EPS - d * d / 4.0) * I;
        intersections.push_back((p[i] + p[j]) / 2.0 + dir);
        intersections.push_back((p[i] + p[j]) / 2.0 - dir);
      }
    }
    
    bool encloses = false;
    for(complex<double> intersection : intersections){
      encloses = true;
      for(int i = 0; i < N; i++){
        if(abs(p[i] - intersection) > R + EPS){
          encloses = false;
        }
      }
      if(encloses) break;
    }
    (encloses ? t : b) = R;
  }
  
  cout << fixed << setprecision(10) << R << endl;
  
  return 0;
}