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 <cstdio>
#include <cmath>
int n;
double x[1005], y[1005], X, Y, d, e;
double dist(double a, double b) {
	return a*a + b*b;
}
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%lf%lf", &x[i], &y[i]);
		X += x[i]; Y += y[i];
	}
	X /= n; Y /= n;
	double P = 0.1;
	for (int i = 0; i < 30000; 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 (d < e) { d = e; f = j; }
		}
		X += (x[f] - X)*P;
		Y += (y[f] - Y)*P;
		P *= 0.999;
	}
	printf("%.6lf", sqrt(d));
}