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 <iostream>
#include <vector>
#include <iomanip>
#include <cmath>

using namespace std;

double dist(double a, double b){
	return a*a + b*b;
}

int main(){
	
	int n; cin >> n;
	double x[n+7], y[n+7];
	double X = 0, Y = 0, d, e;
	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 < 50000; 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(e > d){
				d = e; f = j;
			}
		}
		X += (x[f]-X)*P;
		Y += (y[f]-Y)*P;
		P *= 0.999;
	}

	cout << setprecision(15) << sqrt(d) << '\n';

	return 0;
}