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
#include <bits/stdc++.h>
#define rep(i,n) for (ll i = 0; i < n; ++i)
using namespace std;
using ll = long long;
using ull = long long unsigned;
using P = pair<double,double>;

int n;
vector<P> v(55);


double dist(P a, P b){
	double dx=a.first-b.first;
	double dy=a.second-b.second;
	return sqrt(dx*dx+dy*dy);
}


//(x,y)を中心とする円で全ての頂点を含む時の半径
double f(double x,double y){
	double res=0;
	rep(i,n){
		res=max(res,dist(make_pair(x,y),v[i]));
	}
	return res;
}


//三分探索、x座標固定
double g(double x){
	double l=0, r=1005;
	rep(i,85){
		double c1=(l*2+r)/3;
		double c2=(l+r*2)/3;
		if (f(x,c1)>f(x,c2)) l=c1;else r=c2;
	}
	return f(x,l);

}


int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
cin>>n;
rep(i,n){
	double t1,t2;
	cin>>t1>>t2;
	v[i]=make_pair(t1,t2);
}
double l=0,r=1005;
rep(i,85){
	double c1=(l*2+r)/3;
	double c2=(l+r*2)/3;
	if (g(c1)>g(c2)) l=c1;else r=c2;

}
double ans=g(l);
cout<<setprecision(10)<<ans;

}
