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;

int N;
vector<double> x(0), y(0);

double f(double x0, double y0){
	double ans = 0;
	for(int i = 0; i < N; i++){
		ans = max(ans, hypot(x[i] - x0, y[i] - y0));
	}
//	printf("f(%lf, %lf) = %lf\n", x0, y0, ans);
	return ans;
}

double f(double x0){
	double l = 0, h1, h2, r = 1000;
	for(int i = 0; i < 50; i++){
		h1 = (2 * l + r) / 3;
		h2 = (l + 2 * r) / 3;
		if(f(x0, h1) > f(x0, h2)){
			l = h1;
		}
		else{
			r = h2;
		}
	}
	return f(x0, (l + r) / 2);
}

int main(){
	int i;
	scanf("%d", &N);
	x.assign(N, 0);
	y.assign(N, 0);
	for(i = 0; i < N; i++){
		scanf("%lf%lf", &x[i], &y[i]);
	}
	double l = 0, h1, h2, r = 1000;
	for(i = 0; i < 50; i++){
		h1 = (2 * l + r) / 3;
		h2 = (l + 2 * r) / 3;
		if(f(h1) > f(h2)){
			l = h1;
		}
		else{
			r = h2;
		}
	}
	printf("%.10lf\n", f((l + r) / 2));
	return 0;
}