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;

vector<int> xs, ys;

double dist(double x, double y) {
  double ans = 0;
  for (int i = 0; i < xs.size(); i++) {
    double dx = x - xs.at(i), dy = y - ys.at(i);
    ans = max(ans, sqrt(dx * dx + dy * dy));
  }
  return ans;
}

double calc_y(double x) {
  double y0 = 0, y1 = 1000;
  while (y1 - y0 > 1e-8) {
    double dy = y1 - y0, l = y0 + dy / 3, r = y1 - dy / 3;
    double d1 = dist(x, l);
    double d2 = dist(x, r);
    if (d1 < d2) {
      y1 = r;
    } else {
      y0 = l;
    }
  }
  return (y0 + y1) / 2;
}

int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int xi, yi;
    cin >> xi >> yi;
    xs.push_back(xi);
    ys.push_back(yi);
  }
  double x0 = 0, x1 = 1000;
  while (x1 - x0 > 1e-8) {
    double dx = x1 - x0, l = x0 + dx / 3, r = x1 - dx / 3;
    double d1 = dist(l, calc_y(l));
    double d2 = dist(r, calc_y(r));
    if (d1 < d2) {
      x1 = r;
    } else {
      x0 = l;
    }
  }
  cout << setprecision(17) << dist(x0, calc_y(x0)) << endl;
}