Problem: 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

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <cmath>

#define LMIN 0
#define RMAX 1000
#define REPNUM 80
using namespace std;

typedef struct P
{
  double x, y;
}P;

inline double dist(P a, P b)
{
  double dx = a.x - b.x;
  double dy = a.y - b.y;
  return (dx * dx + dy * dy);
}

double f(vector<P>& p, double x, double y)
{
  double res = 0;
  for(int i = 0; i < p.size(); ++i)
  {
    res = max(res, dist((P){x, y}, p[i]));
  }
  return sqrt(res);
}

double g(vector<P>& p, double x)
{
  double l = LMIN, r = RMAX;
  //for(int i = 0; i < REPNUM; ++i)
  while(abs(l-r) > 1e-7)
  {
    double c1 = (l * 2 + r) / 3;
    double c2 = (l + r * 2) / 3;
    if(f(p, x, c1) > f(p, x, c2))
      l = c1;
    else
      r = c2;
  }
  return f(p, x, l);
}

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

  double l = LMIN, r = RMAX;
  //for(int i = 0; i < REPNUM; ++i)
  while(abs(l-r) > 1e-7)
  {
    double c1 = (l * 2 + r) / 3;
    double c2 = (l + r * 2) / 3;
    if(g(p, c1) > g(p, c2))
      l = c1;
    else
      r = c2;
  }
  printf("%.10f\n", g(p, l));
  return 0;
}