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 n;
double X[55],Y[55];
double dist(double x1,double x2,double y1,double y2){
	return sqrt(double((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)));
}
double dist(double x,double y){
	double rad=0.00;
	for(int i=1;i<=n;i++){
		rad=max(rad,dist(X[i],x,Y[i],y));
	}
	return rad;
}
double func(double x){
	double le=0,ri=1000,mid1,mid2,f1,f2,ans;
	for(int i=1;i<=1000;i++){
		mid1=le+(ri-le)/3;mid2=ri-(ri-le)/3;
		f1=dist(x,mid1);f2=dist(x,mid2);
		if(f1<f2){ans=f1;ri=mid2;}
		else{ans=f2;le=mid1;}
	}
	return ans;
}
int main(){
	#ifdef FILIN
		#ifndef LOCAL
			freopen(FILIN,"r",stdin);
			freopen(FILOUT,"w",stdout);
		#endif
	#endif
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>X[i]>>Y[i];
	} 
	//ternary search
	double le=0,ri=1000,mid1,mid2,f1,f2,ans;
	for(int i=1;i<=1000;i++){
		mid1=le+(ri-le)/3;mid2=ri-(ri-le)/3;
		f1=func(mid1);f2=func(mid2);
		if(f1<f2){ans=f1;ri=mid2;}
		else{ans=f2;le=mid1;}
	}
	cout<<fixed<<setprecision(20)<<ans<<endl;
	return 0;
}