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 int maxn = 50 + 10;
const double eps = 1e-8;
int n;
struct Point
{
	double x, y;
}p[maxn];
double xl, xr, yl, yr;
inline double sqr(const double & x) { return x * x;}
double dis(const Point & a, const Point & b)
{
	return sqr(a.x - b.x) + sqr(a.y - b.y);
}
double f2(double x, double y)
{
	auto Now = Point{x, y};
	double ret = 0;
	for(int i = 1; i <= n; ++i)
		ret = max(ret, dis(Now, p[i]));
	return ret;
}
double f(double x)
{
	yl = 0, yr = 1000;
	while(yr - yl >= eps)
	{
		double lmid = yl + (yr - yl) / 3;
		double rmid = yr - (yr - yl) / 3;
		if(f2(x, lmid) < f2(x, rmid))
			yr = rmid;
		else
			yl = lmid;
	}
	return f2(x, yl);
}
int main()
{
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i)
		scanf("%lf%lf", &p[i].x, &p[i].y);
	xl = 0, xr = 1000;
	while(xr - xl >= eps)
	{
		double lmid = xl + (xr - xl) / 3;
		double rmid = xr - (xr - xl) / 3;
		if(f(lmid) < f(rmid))
			xr = rmid;
		else
			xl = lmid;
	}
	printf("%.7f\n", sqrt(f(xl)));
	return 0;
}