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;
#ifdef LOCAL_DEBUG
  #include "LOCAL_DEBUG.hpp"
#endif
#define int long long

struct P{
  int x, y;
};

signed main(){

  int n; cin >> n;
  vector<P> a(n);
  for(int i = 0; i < n; i++){
    cin >> a[i].x >> a[i].y;
  }

  auto dist = [&](double x, double y){
    double Max = 0;
    for(int i = 0; i < n; i++){
      double dx = x - a[i].x, dy = y - a[i].y;
      Max = max(Max, sqrt(dx*dx + dy*dy));
    }
    return Max;
  };

  auto f = [&](double x){
    double left = -1, right = 1000;
    //[left , t1 , t2 , right]
    for(int lim = 0; lim < 100; lim++){
      double t1 = (left * 2 + right) / 3;
      double t2 = (left + right * 2) / 3;
      (dist(x, t1) > dist(x, t2) ? left = t1 : right = t2);
    }
    return dist(x, left);
  };
  
  double left = -1, right = 1000;
  //[left , t1 , t2 , right]
  for(int lim = 0; lim < 100; lim++){
    double t1 = (left * 2 + right) / 3;
    double t2 = (left + right * 2) / 3;
    (f(t1) > f(t2) ? left = t1 : right = t2);
  }
  printf("%.9f\n", f(left));

  return 0;
}