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++)
#define fi first
#define se second

using namespace std;

typedef long long ll;
typedef pair<double, double> pdd;

int n;
vector<pdd> v;
int iter = 100;

double f(double x, double y){
	double r = 0.0;
	rep(i, n) {
		double a = v[i].fi, b = v[i].se;
		r = max(r, hypot(x - a, y - b));
	}
	return r;
}

double g(double x){
	double left = -1000.0, right = 1000.0;
	rep(i, iter){
		double m1 = (left * 2 + right) / 3;
		double m2 = (left + right * 2) / 3;
		if(f(x, m1) > f(x, m2)) left = m1;
		else right = m2;
	}

	return f(x, left);
}

double solve(){
	double left = -1000.0, right = 1000.0;
	rep(i, iter){
		double m1 = (left * 2 + right) / 3;
		double m2 = (left + right * 2) / 3;
		if(g(m1) > g(m2)) left = m1;
		else right = m2;
	}
	
	return g(left);
}

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> n;

	v.resize(n);
	rep(i, n){
		double x, y;
		cin >> x >> y;
		v[i] = {x, y};
	}
	
	cout << fixed << setprecision(10);
	cout << solve() << endl;
}
