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>
using namespace std;

const double EPS = 1e-9;

class Point
{
public:
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) {}

    double dist(double x1, double y1) const
    {
        double dx = x - x1, dy = y - y1;
        return sqrt(dx * dx + dy * dy);
    }
};

int N;
vector<Point> p;

double eval(double x, double y)
{
    double res = 0;
    for (int i = 0; i < N; i++)
    {
        res = max(res, p.at(i).dist(x, y));
    }
    return res;
}

double eval_x(double x)
{
    double b = 0, u = 1000;
    while (u - b > EPS)
    {
        double mb = (2 * b + u) / 3;
        double mu = (b + 2 * u) / 3;
        if (eval(x, mb) > eval(x, mu))
            b = mb;
        else
            u = mu;
    }
    return eval(x, b);
}

int main()
{
    cout << fixed << setprecision(18);

    cin >> N;
    p = vector<Point>(N);
    for (int i = 0; i < N; i++)
        cin >> p.at(i).x >> p.at(i).y;

    double l = 0, r = 1000;
    while (r - l > EPS)
    {
        double ml = (2 * l + r) / 3;
        double mr = (l + 2 * r) / 3;
        if (eval_x(ml) > eval_x(mr))
            l = ml;
        else
            r = mr;
    }
    cout << eval_x(l) << endl;
    return 0;
}
