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 <iostream>
#include<bits/stdc++.h>
using namespace std;
typedef double ld;
int n,r;
ld a[50],b[50],A,B,l,P,c;
ld dist(ld a,ld b){
    return a*a+b*b;
}
int main(void){
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>a[i]>>b[i];
        A+=a[i];
        B+=b[i];
    }
    A/=n;
    B/=n;
    P=0.1;
    for(int i=0;i<30000;i++){
        l=dist(a[0]-A,b[0]-B);
        r=0;
        for(int j=1;j<n;j++){
            c=dist(a[j]-A,b[j]-B);
            if(l<c){l=c;r=j;}
        }
        A+=(a[r]-A)*P;
        B+=(b[r]-B)*P;
        P*=0.999;
    }
    cout<<setprecision(15)<<sqrt(l)<<endl;
}