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;

struct Point { double x, y; };
int n;
double ans = 10000;
vector<Point> ps;

double dist(Point a, Point b) {
  double dx = a.x-b.x;
  double dy = a.y-b.y;
  return sqrt(dx*dx + dy*dy);
}

double f( double x, double y ) {
    double res = 0;
    for ( int i=0; i<n; i++  ) {
        res = max(res, dist(ps[i], (Point){x,y}));
    }
    return res;
}

double g( double x ) {
    double l=0, r=1000, mi=10000;
    for ( int k=0; k<50; k++ ) {
        double y1 = (l+l+r)/3;
        double y2 = (l+r+r)/3;
        double res1 = f(x, y1);
        double res2 = f(x, y2);
        mi = min({mi, res1, res2});
        if ( res1 > res2 ) l=y1;
        else r=y2;
    }
    ans = min(ans, mi);
    return mi;
}

int main() {
    cin >> n;
    ps.resize(n);
    for ( int i=0; i<n ; i++ ) {
        cin >> ps[i].x >> ps[i].y;
    }
    // 三分探索
    double l=0, r=1000;
    for ( int k=0; k<50; k++ ) {
        double x1 = (l+l+r)/3;
        double x2 = (l+r+r)/3;
        if ( g(x1) > g(x2) ) l=x1;
        else r=x2;
    }
    cout << fixed << setprecision(12) << ans << '\n';
    return 0;
}