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;

using Pos = pair<double, double>;

double dist(const Pos &a, const Pos &b) {
  return hypot(a.first - b.first, a.second - b.second);
}

double max_dist(const Pos &a, vector<Pos> &vp) {
  double res = 0;
  for (auto &p : vp) {
    res = max(res, dist(a, p));
  }

  return res;
}

double min_dist(vector<Pos> &vp, double x) {
  double l = 0, r = 1000;
  for (int i = 0; i < 80; ++i) {
    double c1, c2;
    c1 = (l * 2 + r) / 3;
    c2 = (l + r * 2) / 3;
    if (max_dist({x, c1}, vp) > max_dist({x, c2}, vp)) {
      l = c1;
    } else {
      r = c2;
    }
  }

  return max_dist({x, l}, vp);
}

int main(int argc, const char *argv[]) {
  int n;
  cin >> n;

  vector<Pos> vp(n);
  for (auto &p : vp) {
    cin >> p.first >> p.second;
  }

  double l = 0, r = 1000;
  for (int i = 0; i < 80; ++i) {
    double c1, c2;
    c1 = (l * 2 + r) / 3;
    c2 = (l + r * 2) / 3;
    if (min_dist(vp, c1) > min_dist(vp, c2)) {
      l = c1;
    } else {
      r = c2;
    }
  }

  cout << setprecision(9) << min_dist(vp, l) << '\n';
}