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:
// Use gradient descent
#include <bits/stdc++.h>
#define X first
#define Y second
using namespace std;
typedef pair<double,double> pi;
vector<pi> point;
int n;

double dis(pi A,pi B){
    return sqrt((A.X - B.X)*(A.X - B.X) + (A.Y - B.Y)*(A.Y - B.Y));
}

double f(pi p){
    double res = 0;
    for (pi x: point) res = max(res,dis(p,x));
    return res;
}

int main(){
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n;
    point.resize(n);
    for (int i = 0; i < n; i++) cin >> point[i].X >> point[i].Y;
    
    double ans = 200000;
    double px = 0, py = 0, alpha = 1e2, delta = 1e-7;
    for (int Loop = 0; Loop < 20000; Loop++){
        double dx = (f(pi(px + delta,py)) - f(pi(px - delta,py)))/(2*delta);
        double dy = (f(pi(px,py + delta)) - f(pi(px,py - delta)))/(2*delta);
        px -= alpha*dx;
        py -= alpha*dy;
        alpha *= 0.999;
    }
    cout << fixed << setprecision(8) << f(pi(px,py));

    return 0;
}