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
import java.util.Scanner;

public class Main {
	static int n;
	static int[] x, y;

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		x = new int[n];
		y = new int[n];
		for (int i = 0; i < n; i++) {
			x[i] = sc.nextInt();
			y[i] = sc.nextInt();
		}
		sc.close();

		System.out.println(ternarySearch());
	}

	static double ternarySearch() {
		double gosa = 1e-8;
		double l = 0;
		double r = 1000;
		while (r - l > gosa) {
			double m1 = (l * 2 + r) / 3;
			double m2 = (l + r * 2) / 3;
			double v1 = ternarySearch(m1);
			double v2 = ternarySearch(m2);
			if (v1 <= v2) {
				r = m2;
			} else {
				l = m1;
			}
		}
		return ternarySearch(r);
	}

	static double ternarySearch(double cx) {
		double gosa = 1e-8;
		double l = 0;
		double r = 1000;
		while (r - l > gosa) {
			double m1 = (l * 2 + r) / 3;
			double m2 = (l + r * 2) / 3;
			double v1 = calc(cx, m1);
			double v2 = calc(cx, m2);
			if (v1 <= v2) {
				r = m2;
			} else {
				l = m1;
			}
		}
		return calc(cx, r);
	}

	static double calc(double cx, double cy) {
		double ret = 0;
		for (int i = 0; i < n; i++) {
			double r = Math.hypot(x[i] - cx, y[i] - cy);
			ret = Math.max(ret, r);
		}
		return ret;
	}
}
