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 <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

int N, x[301], y[301];

double get_r(double X, double Y) {
  double ret = 0;

  for (int i = 0; i < N; ++i)
    ret = max(ret, (X - x[i]) * (X - x[i]) + (Y - y[i]) * (Y - y[i]));

  return ret;
}

double searchY(double X) {
  double lo = 0, hi = 1000;

  for (int it = 0; it < 1000; ++it) {
    double a = (2 * lo + hi) / 3;
    double b = (lo + 2 * hi) / 3;

    double ra = get_r(X, a);
    double rb = get_r(X, b);

    if (ra < rb)
      hi = b;
    else
      lo = a;
  }

  return get_r(X, lo);
}

double searchX() {
  double lo = 0, hi = 1000;

  for (int it = 0; it < 1000; ++it) {
    double a = (2 * lo + hi) / 3;
    double b = (lo + 2 * hi) / 3;

    double ra = searchY(a);
    double rb = searchY(b);

    if (ra < rb)
      hi = b;
    else
      lo = a;
  }

  return searchY(lo);
}

int main() {
  scanf("%d", &N);

  for (int i = 0; i < N; ++i) scanf("%d %d", &x[i], &y[i]);

  printf("%.10f\n", sqrt(searchX()));

  return 0;
}