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<iostream>
#include<stdio.h>
#include<math.h>
#include<string>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll long long
const int maxn = 60;
double x[maxn],y[maxn];
int n;
double solve(double xx,double yy)
{
	double ans=0.0;
	for(int i=1;i<=n;i++)
	ans=max(ans,sqrt((x[i]-xx)*(x[i]-xx)+(y[i]-yy)*(y[i]-yy)));
	return ans;
}
double solve(double xx)
{
	double lb=-1e9,ub=1e9;
	double mid,lmid,rmid;
	double ans=1e17;
	while(ub-lb>1e-8)
	{
		lmid=(lb+ub)/2;
		rmid=(lmid+ub)/2;
		if(solve(xx,lmid) <= solve(xx,rmid)) ub = rmid;
        else lb = lmid;
	}
	return solve(xx,lb);
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	scanf("%lf%lf",&x[i],&y[i]);
	double lb=-1e9,ub=1e9;
	double mid,lmid,rmid;
	double ans=1e17;
	while(ub-lb>1e-8)
	{
		lmid=(lb+ub)/2;
		rmid=(lmid+ub)/2;
		if(solve(lmid) <= solve(rmid)) ub = rmid;
        else lb = lmid;
	}
	printf("%.7lf",solve(lb));
	return 0;
}