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(int i = 0; i < (n); ++i)
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int n;
double x[51], y[51];

double f2(double r, double c){
	double ret = 0;
	rep(i,n) chmax(ret, sqrt((x[i]-r)*(x[i]-r) + (y[i]-c)*(y[i]-c)));
	return ret;
}

double f1(double mid){
	double l = 0, r = 1000;
	rep(_,100){
		double mid1 = (l * 2 + r) / 3;
		double mid2 = (l + r * 2) / 3;
		if(f2(mid, mid1) < f2(mid, mid2)) r = mid2;
		else l = mid1;
	}
	return f2(mid, l);
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> n;
	rep(i,n) cin >> x[i] >> y[i];
	double l = 0, r = 1000;
	rep(_,100){
		double mid1 = (l * 2 + r) / 3;
		double mid2 = (l + r * 2) / 3;
		if(f1(mid1) < f1(mid2)) r = mid2;
		else l = mid1;
	}
	double ans = f1(l);
	cout << fixed << setprecision(12) << ans << endl;
	return 0;
}