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>
using namespace std;
#define ll long long
#define lvector vector<ll> 
typedef pair<ll,ll> P;
#define rep(i,n) for(ll (i)=0; i<(n); ++(i))
#define print(a) cout << (a) << endl

double dist(double X,double Y,double x,double y) {
  return sqrt((X-x)*(X-x)+(Y-y)*(Y-y));
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ll n;cin>>n;
  double gx=0,gy=0,d=0,d_=0,pos=0,c=0.1;
  lvector xv(n,0),yv(n,0);
  rep(i,n) {
    cin>>xv[i]>>yv[i];
    gx+=xv[i],gy+=yv[i];
  }
  gx/=n,gy/=n;
  rep(t,30000) {
    d=0;
    for(ll i=0;i<n;++i) {
      d_=dist(gx,gy,xv[i],yv[i]);
      if(d_>d) {d=d_;pos=i;}
    }
    gx-=(gx-xv[pos])*c;
    gy-=(gy-yv[pos])*c;
    c*=0.999;
  }
  d=dist(gx,gy,xv[pos],yv[pos]);
  printf("%.10f",d);
  return 0;
}
