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: Accepted)

#include <bits/stdc++.h>
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define rep(i,N) for(long long i = 0; i < (long long)(N); i++)
#define repr(i,N) for(long long i = (long long)(N) - 1; i >= 0; i--)
#define rep1(i,N) for(long long i = 1; i <= (long long)(N) ; i++)
#define repr1(i,N) for(long long i = (N) ; (long long)(i) > 0 ; i--)
#define each(x,v) for(auto& x : v)
#define all(v) (v).begin(),(v).end()
#define sz(v) ((int)(v).size())
#define ini(...) int __VA_ARGS__; in(__VA_ARGS__)
#define inl(...) long long __VA_ARGS__; in(__VA_ARGS__)
#define ins(...) string __VA_ARGS__; in(__VA_ARGS__)
using namespace std; void solve();
using ll = long long; template<class T = ll> using V = vector<T>;
using vi = V<int>; using vl = V<>; using vvi = V< V<int> >;
constexpr int inf = 1001001001; constexpr ll infLL = (1LL << 61) - 1;
struct IoSetupNya {IoSetupNya() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); cerr << fixed << setprecision(7);} } iosetupnya;
template<typename T, typename U> inline bool amin(T &x, U y) { return (y < x) ? (x = y, true) : false; }
template<typename T, typename U> inline bool amax(T &x, U y) { return (x < y) ? (x = y, true) : false; }
template<typename T, typename U> ostream& operator <<(ostream& os, const pair<T, U> &p) { os << p.first << " " << p.second; return os; }
template<typename T, typename U> istream& operator >>(istream& is, pair<T, U> &p) { is >> p.first >> p.second; return is; }
template<typename T> ostream& operator <<(ostream& os, const vector<T> &v) { int s = (int)v.size(); rep(i,s) os << (i ? " " : "") << v[i]; return os; }
template<typename T> istream& operator >>(istream& is, vector<T> &v) { for(auto &x : v) is >> x; return is; }
void in(){} template <typename T,class... U> void in(T &t,U &...u){ cin >> t; in(u...);}
void out(){cout << "\n";} template <typename T,class... U> void out(const T &t,const U &...u){ cout << t; if(sizeof...(u)) cout << " "; out(u...);}
template<typename T>void die(T x){out(x); exit(0);}
#ifdef NyaanDebug
  #include "NyaanDebug.h"
  #define trc(...) do { cerr << #__VA_ARGS__ << " = "; dbg_out(__VA_ARGS__);} while(0)
  #define trca(v,N) do { cerr << #v << " = "; array_out(v , N);cout << endl;} while(0)
#else
  #define trc(...)
  #define trca(...)
  int main(){solve();}
#endif

using P = pair<ll,ll>; using vp = V<P>;
constexpr int MOD = /**/ 1000000007; //*/ 998244353;
//////////

void solve(){
  ini(N);
  using cd = complex<double>;
  V<complex<double> > v;
  rep(i , N){
    double x,y; in(x,y); v.eb(x , y);
  }

  auto f = [&](double x,double y) -> double{
    double ret = 0.0;
    each(p , v) amax(ret , abs(p - cd(x  ,y)));
    return ret;
  };

  auto g = [&](double x) -> double{
    // yを三分探索
    double l = -2e3 , r = 2e3;
    rep(_ , 100){
      double m1 = (l * 2 + r) / 3;
      double m2 = (l + r * 2) / 3;
      if(f(x , m1) < f(x , m2)) r = m2;
      else l = m1;
    }
    return f (x , (l + r) / 2.0);
  };

  // yを三分探索
  double l = -2e3 , r = 2e3;
  rep(_ , 100){
    double m1 = (l * 2 + r) / 3;
    double m2 = (l + r * 2) / 3;
    if(g(m1) < g(m2)) r = m2;
    else l = m1;
  }
  out(g((l + r) / 2));

}