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:
// Code by H~$~C
#include <bits/stdc++.h>
using namespace std;

static const int Maxn = 55;
static const double EPS = 1e-9;

int n;

struct point {
  double x, y;
} a[Maxn];
inline double get_dist(double x, double y) {
  double mx = 0;
  for (int i = 1; i <= n; ++i) {
    double now = sqrt((x - a[i].x) * (x - a[i].x) + (y - a[i].y) * (y - a[i].y));
    mx = max(mx, now);
  }
  return mx;
}

double f(double x) {
  double l = 0, r = 1000;
  while (r - l > EPS) {
    double lmid = l + (r - l) / 3.0;
    double rmid = r - (r - l) / 3.0;
    if (get_dist(x, lmid) > get_dist(x, rmid)) l = lmid;
    else r = rmid;
  }
  return get_dist(x, l);
}

int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i)
    scanf("%lf%lf", &a[i].x, &a[i].y);
  double l = 0, r = 1000;
  while (r - l > EPS) {
    double lmid = l + (r - l) / 3.0;
    double rmid = r - (r - l) / 3.0;
    if (f(lmid) > f(rmid)) l = lmid;
    else r = rmid;
  }
  printf("%.7lf\n", f(l));
  return 0;
}