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;
int X[55],Y[55];
double dist(double x1,double y1,double x2,double y2){
    return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));
}
double max_dist(double x,double y,int n){
    double ans=0;
	for(int i=1;i<=n;i++) 
		ans=max(ans,dist(x,y,X[i],Y[i]));
    return ans;
}
double func(double x,int n){
    double le=0,ri=1000,ans;
    for(int i=1;i<=1000;i++){
    	double mid1=le+(ri-le)/3;
        double f1=max_dist(x,mid1,n);
        double mid2=ri-(ri-le)/3;
        double f2=max_dist(x,mid2,n);
        if(f1<f2){
        	ri=mid2;
        	ans=f1;
		}
		else{
			le=mid1;
			ans=f2;
		}
    }
    return ans;
}
int n;
void solve(){
	cin>>n;
	for(int i=1;i<=n;i++) cin>>X[i]>>Y[i];
	double le=0,ri=1000,ans;
    for(int i=1;i<=1000;i++){
        double mid1=le+(ri-le)/3;
        double mid2=ri-(ri-le)/3;
        double f1=func(mid1,n);
        double f2=func(mid2,n);
        if(f1<f2){
        	ri=mid2;
        	ans=f1;
		}
		else{
			le=mid1;
			ans=f2;
		}
    }
    cout<<fixed<<setprecision(50)<<ans<<endl;
}
int main(){
	solve();
	return 0;
}