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;
using point = complex<double>;
const double EPS = 1e-9;

int n;
vector<point> pts, center;

bool check(double R) {
  center.clear();
  for (int i = 1; i < n; ++i) {
    for (int j = 0; j < i; ++j) {
      double d = abs(pts[i] - pts[j]);
      double h = R * R - d * d / 4.0;
      if (h < 0) continue;
      h = sqrt(h);
      point C = (pts[i] + pts[j]) / 2.0;
      point vec = point(h / d * (pts[j].imag() - pts[i].imag()), -h / d * (pts[j].real() - pts[i].real()));
      center.push_back(C + vec);
      center.push_back(C - vec);
    }
  }
  for (auto &p: center) {
    bool ok = true;
    for (auto &q: pts) {
      if (abs(p - q) - EPS > R) {
        ok = false;
        break;
      }
    }
    if (ok) return true;
  }
  return false;
}

int main() {
  ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
  cin >> n;
  for (int i = 1, x, y; i <= n; ++i) {
    cin >> x >> y;
    pts.emplace_back(x, y);
  }
  check(0.5);
  double lt = 0, rt = 1000;
  for (int i = 1; i <= 200; ++i) {
    double mid = (lt + rt) / 2.0;
    if (check(mid)) rt = mid;
    else lt = mid;
  }
  cout << fixed << setprecision(12) << lt;
  return 0;
}