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
//verma_ankit484

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

typedef long long ll;
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define pb push_back
const ll mod = (ll) 1e9+7;

using namespace std;
//using namespace __gnu_pbds;

//#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update>


double getdist(double p, double q) {
  return p*p+q*q;
}

int main() {
  ll n;
  cin >> n;
  double X, Y;
  vector<pair<ll, ll>> vc(n);
  for (ll i = 0; i < n; i++) {
    cin >> vc[i].first >> vc[i].second;
    X += vc[i].first, Y += vc[i].second;
  }
  X /= n, Y /= n;
  double P = 0.1, d, e;
  for (ll i = 0; i < 20000; i++) {
    ll f = 0;
    d = getdist(X - vc[0].first, Y - vc[0].second);
    for (ll j = 1; j < n; j++) {
      e = getdist(X - vc[j].first, Y - vc[j].second);
      if (d < e) { d = e; f = j; }
    }
    X += (vc[f].first - X)*P;
    Y += (vc[f].second - Y)*P;
    P *= 0.999;
  }
  cout << fixed << setprecision(10) << sqrt(d) << endl;
  return 0;
}
