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 <iomanip>
#include <cmath>
#include <functional>
#include <iostream>
using namespace std;

int main() {
    int n; cin >> n;
    int x[n], y[n];
    for (int i = 0; i < n; i++) cin >> x[i] >> y[i];
    function<double(double, double)> f = [&](double p, double q) {
        double maxd2 = 0;
        for (int i = 0; i < n; i++) {
            double dx = p - x[i], dy = q - y[i];
            maxd2 = max(maxd2, dx * dx + dy * dy);
        }
        return sqrt(maxd2);
    };
    function<double(double)> g = [&](double p) {
        double ly = -2000, ry = 2000, ans;
        for (int j = 0; j < 100; j++) {
            double y1 = (2 * ly + ry) / 3, y2 = (ly + 2 * ry) / 3,
                f1 = f(p, y1), f2 = f(p, y2);
            if (f1 < f2) ry = y2; else ly = y1;
            ans = f1;
        }
        return ans;
    };
    double lx = -2000, rx = 2000, ans;
    for (int i = 0; i < 100; i++) {
        double x1 = (2 * lx + rx) / 3, x2 = (lx + 2 * rx) / 3,
            g1 = g(x1), g2 = g(x2);
        if (g1 < g2) rx = x2; else lx = x1;
        ans = g1;
    }
    cout << setprecision(12) << ans << endl;
}
