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>
using namespace std;
typedef long long ll;
#define rep(i,n) for (ll i = 0; i < (n); ++i)
#define vcout(v) rep(i,v.size()) cout << v[i] << " "; cout << endl
#define mcout(m) rep(j,m.size()){vcout(m[j]);}

struct P{double x; double y;};

// 2点間の距離
double dist(P p1, P p2){
	double dx = p1.x - p2.x;
	double dy = p1.y - p2.y;
	return sqrt(dx*dx + dy*dy);
}

void mainsolve(){

	ll n;
	cin >> n;
	vector<P> ps(n);
	rep(i,n){
		double x,y;
		cin >> x >> y;
		ps[i] = P{x,y};
	}

	// (x,y)を中心としたときの最遠点との距離
	auto f = [&](double x, double y){
		P center = P{x,y};
		double res = 0;
		rep(i,n){
			res = max(res, dist(center, ps[i]));
		}
		return res;
	};

	// fにおいてxを固定したときのfの最小値
	auto g = [&](double x){
		double l = 0;
		double r = 1000;
		rep(i,100){
			double p1 = (l*2+r)/3;
			double p2 = (l+r*2)/3;
			if(f(x,p1) > f(x,p2)) l = p1;
			else r = p2;
		}
		return f(x,l);
	};

	double l = 0;
	double r = 1000;
	rep(i,100){
		double p1 = (l*2+r)/3;
		double p2 = (l+r*2)/3;
		if(g(p1) > g(p2)) l = p1;
		else r = p2;
	}

	printf("%.10f\n", g(l));
	return;
}

int main(){
	
	//ll TC;cin>>TC;rep(I,TC){cout<<"case:"<<I+1<<"---"<<endl; // comment out!
		mainsolve();
	//} // comment out!
	return 0;
}