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>
using namespace std;
vector<pair<double,double>>pts(100);
double distance(double x, double y,int n){
    double ans = 0;
    for(int i=0; i<n; i++){
        double d=sqrt((pts[i].first-x)*(pts[i].first-x) + (pts[i].second-y)*(pts[i].second-y));
        ans=max(ans, d);
    }
 return ans;
}

double search(double x,int n){
    double l=0,r=1000;
    int t=100;
     while(t--){
        double mid1=(2*l+r)/3.0;
        double mid2=(l+2*r)/3.0;

        if(distance(x,mid1,n)>distance(x,mid2,n)) l=mid1;
        else r=mid2;
    }
    return distance(x,l,n);
}

int main(){
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
        int x,y;
        cin>>x>>y;
        pts[i].first=x;
        pts[i].second=y;
    }

    double l=0,r=1000;
    for(int i=0;i<100;i++){
        double mid1=(2*l+r)/3.0;
        double mid2=(l+2*r)/3.0;

        if(search(mid1,n)>search(mid2,n)) l=mid1;
        else r=mid2;
    }
    
    cout<<fixed<<setprecision(20)<<search(l,n)<<endl;
    return 0;
}