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;
#define rep(i,n) for(int i=0; i<(n); i++)
int n;
double x[55], y[55], ox, oy, d, EPS = 1e-7, ans = 2000;
inline double dis(double x, double y){return x*x+y*y;};
int main(){
	scanf("%d", &n);
	rep(i,n) scanf("%lf%lf", x+i, y+i);
	rep(i,n) rep(j,i){
		bool ok = true;
		ox = (x[i]+x[j]) / 2;
		oy = (y[i]+y[j]) / 2;
		d = dis(ox-x[i],oy-y[i]);
		rep(k,n) if(d + EPS < dis(ox-x[k],oy-y[k])) ok = false;
		if(ok) ans = min(ans, sqrt(d));
		else{
			rep(k,j){
				bool ok = true;
				double x1 = x[k]-x[i], x2 = x[k]-x[j], y1 = y[k]-y[i], y2 = y[k]-y[j], x3 = (x[k]+x[i])/2, x4 = (x[k]+x[j])/2, y3 = (y[k]+y[i])/2, y4 = (y[k]+y[j])/2;
				if(abs(x1*y2 - x2*y1) < EPS) continue;
				double t = ((x3 - x4) * x2 + (y3 - y4) * y2) / (x1*y2 - x2*y1);
				ox = t * y1 + x3;
				oy = -t * x1 + y3;
				d = dis(ox-x[i],oy-y[i]);
				rep(l,n) if(d + EPS < dis(ox-x[l],oy-y[l])) ok = false;
				if(ok) ans = min(ans, sqrt(d));
			}
		}
	}
	printf("%.9lf\n", ans);
}