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 <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <cstdio>
using namespace std;

int main()
{
    int N;
    cin>>N;
    vector<double> x(N), y(N);
    for (int i=0; i<N; i++)
        cin>>x[i]>>y[i];

    auto f1 = [&](double ox, double oy)
    {
        double r = 0;
        for (int i=0; i<N; i++)
            r = max(r, hypot(x[i]-ox, y[i]-oy));
        return r;
    };

    auto f2 = [&](double ox)
    {
        double l = 0;
        double r = 1000;
        for (int i=0; i<100; i++)
        {
            double m1 = (2*l+r)/3;
            double m2 = (l+2*r)/3;
            if (f1(ox, m1) < f1(ox, m2))
                r = m2;
            else
                l = m1;
        }
        return f1(ox, l);
    };

    auto f3 = [&]()
    {
        double l = 0;
        double r = 1000;
        for (int i=0; i<100; i++)
        {
            double m1 = (2*l+r)/3;
            double m2 = (l+2*r)/3;
            if (f2(m1) < f2(m2))
                r = m2;
            else
                l = m1;
        }
        return f2(l);
    };

    printf("%.10f\n", f3());
}
