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>
#define lque queue<ll>
#define lpque priority_queue<ll>
#define dlpque priority_queue<ll,lvector,greater<ll>>
#define P pair<ll,ll>
#define ALL(a) (a).begin(),(a).end()
#define rep(i,n) for(ll (i)=0; (i)<(n); ++(i))
#define print(a) cout << (a) << endl

double dist(double gx,double gy, P p) {
  return sqrt((gx-p.first)*(gx-p.first)+(gy-p.second)*(gy-p.second));
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll n,x,y;cin>>n;
  double gx=0.0,gy=0.0,r=0.0,r_=0.0,c=0.5;
  vector<P> points(n);
  rep(i,n) {cin>>x>>y;gx+=x;gy+=y;points[i]=P(x,y);}
  gx/=n,gy/=n;
  P far;
  rep(i,1e6) {
    r=0.0;
    for(P p:points) {
      r_=dist(gx,gy,p);
      if(r<r_) {r=r_;far=p;} 
    }
    gx+=c*(far.first-gx);
    gy+=c*(far.second-gy);
    c*=0.999;
  }
  for(P p:points) r=max(r,dist(gx,gy,p));
  printf("%.10f\n",r);
  return 0;
}