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 <bits/stdc++.h>
#define fi first
#define se second
#define m_p make_pair
#define p_b push_back
#define e_b emplace_back
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
using namespace std;
using ll = long long;
using ld = long double;
template<class T> bool chmax(T &a,T b){if(a<b){a=b;return true;}return false;}
template<class T> bool chmin(T &a,T b){if(a>b){a=b;return true;}return false;}

int N;
vector<ld> x,y;

ld calc(ld X,ld Y){
  ld ans=0;
  for(int i=0;i<N;i++){
    ld tmpx=abs(X-x.at(i));
    ld tmpy=abs(Y-y.at(i));
    chmax(ans,sqrt(tmpx*tmpx + tmpy*tmpy));
  }
  return ans;
}

ld f(ld X){
  ld l=0;
  ld r=1000;
  for(int i=0;i<100;i++){
    ld midl=(2*l+r)/3;
    ld midr=(l+2*r)/3;
    if(calc(X,midl)>calc(X,midr)) l=midl;
    else  r=midr;
  }
  return calc(X,l);
}

int main(){
  ios_base::sync_with_stdio(false);cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> N;
  x.resize(N);
  y.resize(N);
  for(int i=0;i<N;i++)  cin >> x.at(i) >> y.at(i);

  ld l=0;
  ld r=1000;
  for(int i=0;i<100;i++){
    ld midl=(2*l+r)/3;
    ld midr=(l+2*r)/3;
    if(f(midl)>f(midr))  l=midl;
    else  r=midr;
  }
  cout << f(l) << endl;

  return 0;
}