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 <bits/stdc++.h>
//1000000000000223
#define ll long long
#define ld long double
#define IO ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
int n ;
ld x[60], y[60], ans;
ld solve(ld xx, ld yy)
{
    ld ans=0;
    for(int i=1; i<=n; i++)
        ans=max(ans,hypot(xx-x[i],yy-y[i]));
    return ans;
}
ld solve2(ld xx)
{
    ld low=-1e3, high=1e3, m1, m2, cur;
    for(int i=1; i<=70; i++)
    {
        m1=low+(high-low)/3;
        m2=high-(high-low)/3;
        ld ans1=solve(xx,m1), ans2=solve(xx,m2);
        cur=min(ans1,ans2);
        if(ans1<ans2)
            high=m2;
        else
            low=m1;
    }
    return cur;
}
ld solve1()
{
    ld low=-1e3, high=1e3, m1, m2;
    for(int i=1; i<=70; i++)
    {
        m1=low+(high-low)/3;
        m2=high-(high-low)/3;
        ld ans1=solve2(m1), ans2=solve2(m2);
        ans=min(ans1,ans2);
        if(ans1<ans2)
            high=m2;
        else
            low=m1;
    }
    return ans;
}
int main()
{
    IO
    cin>>n;
    for(int i=1; i<=n; i++)
        cin>>x[i]>>y[i];
    cout<<fixed<<setprecision(10)<<solve1();
    return 0;
}
