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=1e5+5;
struct Point{
	double x,y;
}p[N],rt;
double R;
double sqr(double x) {return x*x;}
double dis(Point x,Point y) {return sqrt((x.x-y.x)*(x.x-y.x)+(x.y-y.y)*(x.y-y.y));}
bool incircle(Point x) {return dis(rt,x)<=R;}
Point solve(double a,double p,double c,double d,double e,double f) {
	double y=(f*a-c*d)/(p*d-e*a);
	double x=(f*p-c*e)/(a*e-p*d);
	return {x,y};
}
int main() {
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++) {
		scanf("%lf%lf",&p[i].x,&p[i].y);
	}
	R=0;
	for(int i=1;i<=n;i++) {
		if(incircle(p[i])) continue;
		rt=p[i];
		R=0;
		for(int j=1;j<i;j++) {
			if(incircle(p[j])) continue;
			rt.x=(p[i].x+p[j].x)/2;
			rt.y=(p[i].y+p[j].y)/2;
			R=dis(rt,p[i]);
			for(int k=1;k<j;k++) {
				if(incircle(p[k])) continue;
				rt=solve(
					p[i].x-p[j].x,p[i].y-p[j].y,(sqr(p[j].x)+sqr(p[j].y)-sqr(p[i].x)-sqr(p[i].y))/2,
					p[i].x-p[k].x,p[i].y-p[k].y,(sqr(p[k].x)+sqr(p[k].y)-sqr(p[i].x)-sqr(p[i].y))/2);
				R=dis(rt,p[i]);
			}
		}
	}
	printf("%.12f\n",R);
	return 0;
}