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 <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <math.h>
#include <cassert>
#define rep(i,n) for(int i = 0; i < n; ++i )
using namespace std;
using ll = long long;
using P = pair<int,int>;

int n;
double x[50],y[50];

double dist(double s, double t) {
  double ans = 0;
  rep(i,n){
    ans = max(ans,sqrt((s-x[i])*(s-x[i])+(t-y[i])*(t-y[i])));
  }
  return ans;
}

int main() {
  cin >> n;
  rep(i,n){
    cin >> x[i] >> y[i];
  }

  auto f = [](double xi){
    double yl = 0,yr = 1000;
    rep(i,100) {
      double cl = (yl*2+yr)/3, cr = (yl+yr*2)/3;
      if (dist(xi,cl) > dist(xi,cr)) {
        yl = cl;
      } else {
        yr = cr;
      }
    }
    return dist(xi,yl);
  };
  double xl = 0,xr = 1000;
  rep(i,100) {
    double cl = (xl*2+xr)/3, cr = (xl+xr*2)/3;
    if (f(cl) > f(cr)) {
      xl = cl;
    } else {
      xr = cr;
    }
  }
  printf("%.10f\n",f(xl));
}
