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 unsigned long long ull;
typedef long long ll;

#define EPS (double)(1e-7)

int n;
double x[50], y[50];

double dist(double x1, double y1, double x2, double y2){
	return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

double find(){
	double cx=0.0, cy=0.0, tx=0.0, ty=0.0;
	double mvrate=1.0, maxdist;
	bool flg=false;
	while(!flg){
		for(int p=0; p<100; p++){
			int k;
			maxdist=0.0;
			for(int i=0; i<n; i++){
				double d=dist(cx, cy, x[i], y[i]);
				if(maxdist<d){
					maxdist=d;
					k=i;
				}
			}
			cx+=(x[k]-cx)*mvrate;
			cy+=(y[k]-cy)*mvrate;
			if(dist(tx, ty, cx, cy)<EPS){
				flg=true;
				break;
			}
			tx=cx;
			ty=cy;
		}
		mvrate/=2.0;
	}
	return maxdist;
}

int main(){
	cin >> n;
	for(int i=0; i<n; i++) cin >> x[i] >> y[i];
	printf("%.9f\n", find());
	return 0;
}