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;
#define rep(i,N) for(int i=0;i<int(N);++i)
#define rep1(i,N) for(int i=1;i<int(N);++i)
#define all(a) (a).begin(),(a).end()
#define print(v) { cerr<<#v<<": [ "; for(auto _ : v) cerr<<_<<", "; cerr<<"]"<<endl; }
#define printpair(v) { cerr<<#v<<": [ "; for(auto _ : v) cerr<<"{"<<_.first<<","<<_.second<<"}"<<", "; cerr<<"]"<<endl; }
#define bit(k) (1LL<<(k))
typedef long long ll;

int n;
double x[1005], y[1005], X, Y, d, e;
double dist(double a, double b) {
	return a*a + b*b;
}
int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);

	cin>>n;
	for (int i = 0; i < n; i++) {
		cin>>x[i]>>y[i];
		X += x[i]; Y += y[i];
	}
	X /= n; Y /= n;
	double P = 0.1;
	for (int i = 0; i < 30000; i++) {
		int f = 0;
		d = dist(X - x[0], Y - y[0]);
		for (int j = 1; j < n; j++) {
			e = dist(X - x[j], Y - y[j]);
			if (d < e) { d = e; f = j; }
		}
		X += (x[f] - X)*P;
		Y += (y[f] - Y)*P;
		P *= 0.999;
	}
    cout<<sqrt(d)<<endl;
	//printf("%.3lf %.3lf\n%.3lf", X, Y, sqrt(d));
}