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>
typedef long long int ll;
using namespace std;
#define X first
#define Y second
#define pb push_back
#define fastio ios_base::sync_with_stdio(false);cout.tie(0);cin.tie(0);

const ll mod=1e9+7;
const ll MAXN=1e6+5;
const ll inf=1e18;



int main()
{
    fastio;
    cout<<fixed<<setprecision(15);
    ll n;
    cin>>n;
    vector<ll > x(n),y(n);
    for(int i=0;i<n;i++) cin>>x[i]>>y[i];
    double mi=inf-.0001;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            bool ok=true;
            for(int p=0;p<n;p++){
                ll a=x[p]-x[i],b=x[p]-x[j],c=y[p]-y[i],d=y[p]-y[j];
                if(a*b+c*d>0) ok=false;
            }
            if(ok){
                double a=x[j]-x[i],b=y[j]-y[i];
                a=a*a+b*b;
                a=sqrt(a);
                a=a/2;
                mi=min(mi,a);
            }
        }
    }
    cout<<mi<<endl;
}