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 <bits/stdc++.h>
using namespace std;

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<double> x(n), y(n);
  for (int i = 0; i < n; ++i) {
    cin >> x[i] >> y[i];
  }
  auto d = [&](double cx, double cy) {
    double res = 0;
    for (int i = 0; i < n; ++i) {
      res = max(res, hypot(x[i] - cx, y[i] - cy));
    }
    return res;
  };
  auto fn = [&](double a) {
    double ly = 0, ry = 1000;
    for (int _ = 100; _--; ) {
      double mly = (2 * ly + ry) / 3, mry = (ly + 2 * ry) / 3;
      if (d(a, mly) < d(a, mry)) {
        ry = mry;
      } else {
        ly = mly;
      }
    }
    return d(a, ly);
  };
  double lx = 0, rx = 1000;
  for (int _ = 100; _--; ) {
    double mlx = (2 * lx + rx) / 3, mrx = (lx + 2 * rx) / 3;
    if (fn(mlx) < fn(mrx)) {
      rx = mrx;
    } else {
      lx = mlx;
    }
  }
  cout << fixed << setprecision(20);
  cout << fn(lx) << '\n';
}
