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)
using namespace std;
typedef long long ll;
struct P {
  double x, y;
};

int n;
vector<P> p;

double dist(P a, P b) {
  double dx = a.x - b.x;
  double dy = a.y - b.y;
  return sqrt(dx * dx + dy * dy);
}

// 最大半径
double f(double x, double y) {
  double res = 0;
  rep(i, n) {
    res = max(res, dist((P){x, y}, p[i]));
  }
  return res;
}

// x座標を決めた時のy方向の探索
double g(double x) {
  double l = 0, r = 1000;
  rep(i, 70) {
    double c1 = (l * 2 + r) / 3;
    double c2 = (l + r * 2) / 3;
    if (f(x, c1) > f(x, c2)) {
      l = c1;
    } else {
      r = c2;
    }
  }
  return f(x, l);
}

int main() {
  cin >> n;
  p = vector<P>(n);
  rep(i, n) cin >> p[i].x >> p[i].y;
  double l = 0, r = 1000;
  rep(i, 70) {
    double c1 = (l * 2 + r) / 3;
    double c2 = (l + r * 2) / 3;
    if (g(c1) > g(c2)) {
      l = c1;
    } else {
      r = c2;
    }
  }

  double ans = g(l);
  printf("%.10f\n", ans);
  return 0;
}
