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>

#define rep(i, n) for(int (i) = 0; i < (n); ++(i))

typedef long long ll;
const ll MOD = 1e9 + 7;

struct Point {
  double x, y;
};

double dist(const Point &a, const Point &b) {
  double dx = a.x - b.x;
  double dy = a.y - b.y;
  return std::sqrt(dx * dx + dy * dy);
}

int main() {
  int N;
  std::cin >> N;

  std::vector<Point> ps(N);
  rep(i, N) {
    std::cin >> ps[i].x >> ps[i].y;
  }

  auto f = [&](double x, double y) {
    double ret = 0;
    rep(i, N) {
      ret = std::max(ret, dist(Point{x, y}, ps[i]));
    }
    return ret;
  };

  const double EPS = 1e-10;
  auto g = [&](double x) {
    double l = 0;
    double r = 1000;
    while(r - l > EPS) {
      double ml = (l*2 + r) / 3;
      double mr = (l + r*2) / 3;
      if(f(x, ml) > f(x, mr)) {
        l = ml;
      } else {
        r = mr;
      }
    }
    return f(x, l);
  };

  double l = 0;
  double r = 1000;
  while(r - l > EPS) {
    double ml = (l*2 + r) / 3;
    double mr = (l + r*2) / 3;
    if(g(ml) > g(mr)) {
      l = ml;
    } else {
      r = mr;
    }
  };

  std::cout << std::fixed << std::setprecision(10) << g(l) << '\n';
}