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;

int n;
struct P {
	double x, y;
};
vector<P> p;

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

double f(double x, double y){
	double res = 0;
	for(int i = 0; i < n; i++){
		res = max(res, dist((P){x, y}, p[i]));
	}
	return res;
}

double g(double x){
	double l = 0, r = 1000;
	for(int i = 0; i < 100; i++){
		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() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> n;
	p = vector<P>(n);
	for(int i = 0; i < n; i++){
		cin >> p[i].x >> p[i].y;
	}
	double l = 0, r = 1000;
	for(int i = 0; i < 100; i++){
		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 << fixed << setprecision(10) << ans << endl;
	return 0;
}