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:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int n;
double x[55], y[55], centerX, centerY, maxD, currD;

double dist(double a, double b){
  return a*a + b*b;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
  cin>>n;
  for(int i=0; i<n; ++i){
    cin>>x[i]>>y[i];
    centerX += x[i];
    centerY += y[i];
  }
  centerX /= n;
  centerY /= n;
  double P = 0.1;
  for(int i=0; i<70000; ++i){
    maxD = dist(centerX - x[0], centerY - y[0]);
    int at = 0;
    for(int j=1; j<n; ++j){
      currD = dist(centerX - x[j], centerY - y[j]);
      if( currD > maxD){
        maxD = currD;
        at = j;
      }
    }
    centerX += (x[at]-centerX) * P;
    centerY += (y[at]-centerY) * P;
    P *= 0.998;
  }
  cout.precision(10);
  cout<<fixed<<sqrt(maxD)<<'\n';

	return 0;
}