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<bits/stdc++.h>
using namespace std;
const int N=55;
struct point
{
	double x,y;
}a[N],k;
inline double dis(point a,point b){return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}
int main()
{
	int n,num=100000,p;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lf%lf",&a[i].x,&a[i].y);
		k.x+=a[i].x,k.y+=a[i].y;
	}
	k.x/=n,k.y/=n;
	double zh=1.0/n;
	while(num--)
	{
		p=1;
		for(int i=2;i<=n;i++)
			if(dis(k,a[p])<dis(k,a[i]))
				p=i;
		k.x+=(a[p].x-k.x)*zh;
		k.y+=(a[p].y-k.y)*zh;
		zh*=0.999;
	}
	printf("%.12lf\n",sqrt(dis(k,a[p])));
	return 0;
}