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;
        #define int long long


        signed main() {
         int n;
         cin>>n;
         long double x[n],y[n],ans=0,r,a,b,c;
        for(int i=0;i<n;i++)cin>>x[i]>>y[i];
        if(n==2){
            cout<< std::setprecision(20)<<sqrt((x[0]-x[1])*(x[0]-x[1])+(y[0]-y[1])*(y[0]-y[1]))/2;
            return 0;
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<i;j++){
                for(int k=0;k<j;k++){
                   a=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
                   b=sqrt((x[k]-x[j])*(x[k]-x[j])+(y[k]-y[j])*(y[k]-y[j]));
                   c=sqrt((x[k]-x[i])*(x[k]-x[i])+(y[k]-y[i])*(y[k]-y[i]));
                   if(a*a<b*b+c*c&&b*b<a*a+c*c&&c*c<a*a+b*b)r=2*a*b*c/sqrt((a+b+c)*(-a+b+c)*(a-b+c)*(a+b-c));
                   else r=max(a,max(b,c));
                   //r=min(max(a,max(b,c)),2*a*b*c/sqrt((a+b+c)*(-a+b+c)*(a-b+c)*(a+b-c)));
                   ans=max(ans,r/2);
                }
            }
        }
        cout << std::setprecision(20) <<ans;
          
            
         return 0;

        }