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:
// Chat tam phan
#include <bits/stdc++.h>
using namespace std;
vector<pair<double,double>> p;
int n;


double dis(pair<double,double> u,pair<double,double> v){
    return sqrt((u.first - v.first)*(u.first - v.first) + (u.second - v.second)*(u.second - v.second));
}

double f(double x,double y){
    double ans = 0;
    for (auto pp : p) ans = max(ans,dis(pp,make_pair(x,y)));
    return ans;
}


double g(double y){
    double lef = 0, rig = 1000;
    for (int Loop = 0; Loop < 1000; Loop++){
        double x1 = (lef*2 + rig)/3;
        double x2 = (lef + 2*rig)/3;
        if (f(x1,y) < f(x2,y)) rig = x2;
        else lef = x1;
    }
    return f(lef,y);
}

int main(){
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n;
    p.resize(n);
    for (int i = 0; i < n; i++) cin >> p[i].first >> p[i].second;

    double lef = 0, rig = 1000;
    for (int Loop = 0; Loop < 100; Loop++){
        double y1 = (2*lef + rig)/3;
        double y2 = (lef + 2*rig)/3;
        if (g(y1) < g(y2)) rig = y2;
        else lef = y1;
    }
    cout << fixed << setprecision(10) << g(lef);
    return 0;
}