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<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;
#define INF 0x3f3f3f3f
const int maxn=60;
#define _rep(i,a,b) for(int i=(a);i<=(b);++i)
int n,a[maxn][2];
double EPS=1e-10;

double dist(int x1,int y1,double x2,double y2)
{
    double lenth=(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
    return sqrt(lenth);
}
double f(double x,double y)
{
    double ans=0;
    for(int i=1;i<=n;i++)
        ans=max(ans,dist(a[i][0],a[i][1],x,y));
    return ans;
}
double checky(double x)
{
    double l=-1000,r=1000;
    while(r-l>EPS)    
    {
        double y1=(l*2+r)/3,y2=(l+2*r)/3;
        if(f(x,y1)>f(x,y2))
            l=y1;
        else
            r=y2;
    }
    // cout<<x<<" "<<l<<endl;
    return f(x,l);
}  

int main()
{
    #ifdef __LOCAL__
        freopen("1.in","r",stdin);
        freopen("1.out","w",stdout);
    #endif
    cin>>n;
    _rep(i,1,n)
        cin>>a[i][0]>>a[i][1];
    double l=-1000,r=1000;
    while(r-l>EPS)
    {
        double x1=(l*2+r)/3,x2=(l+2*r)/3;
        double f1=checky(x1),f2=checky(x2);
        if(f1>f2)
            l=x1;
        else
            r=x2;
    }
    printf("%.18lf\n",checky(l));
    return 0;
}