Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

2
0 0
1 0

Sample Output 1

0.500000000000000000

Both points are contained in the circle centered at (0.5,0) with a radius of 0.5.

Sample Input 2

3
0 0
0 1
1 0

Sample Output 2

0.707106781186497524

Sample Input 3

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

Sample Output 3

6.726812023536805158

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

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define repp(i, a, b) for (int i = a; i <= (b); ++i)
#define repr(i, a, b) for (int i = a; i >= (b); --i)
#define bit(n) (1LL << (n))
#define sz(x) ((ll)(x).size())
typedef long long ll;
const int INF = 1001001001;
const ll LINF = 1001001001001001001ll;
const int MOD = 1000000007;

template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}

template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}


double dist(double x, double y, double xx, double yy) {
  return sqrt(pow(x - xx, 2) + pow(y - yy, 2));
}

double EPS = 1.0e-10;

int main() {
  int N;
  cin >> N;
  vector<double> x(N), y(N);
  rep(i, N) {
    cin >> x[i] >> y[i];
  }


  double ans = 100000.0;

  repp(i, 0, N - 1) {
    repp(j, i + 1, N - 1) {
      double X = (x[i] + x[j]) / 2.0, Y = (y[i] + y[j]) / 2.0;
      double r = dist(x[i], y[i], x[j], y[j]) / 2.0;

      bool ok = true;
      rep(k, N) {
        if (k == i || k == j)
          continue;
        if (dist(X, Y, x[k], y[k]) > r + EPS)
          ok = false;
      }
      if (ok) {
        chmin(ans, r);
      }
    }
  }
  printf("%.14f\n", ans);  // Lf?
}