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 <iostream>
#include <vector>
#include <climits>
#include <iomanip>
#include <cmath>
using namespace std;
using uint = unsigned int;

int main() {
	
	uint N;
	cin >> N;
	
	vector<double> X(N);
	vector<double> Y(N);
	
	double lx = 1000;
	double hx = 0;
	double ly = 1000;
	double hy = 0;
	
	for(uint i=0; i<N; i++) {
		cin >> X[i] >> Y[i];
		lx = min(lx, X[i]);
		hx = max(hx, X[i]);
		ly = min(ly, Y[i]);
		hy = max(hy, Y[i]);
	}
	
	double curX = (hx + lx) / 2;
	double curY = (hy + ly) / 2;
	double nxtX = 0;
	double nxtY = 0;
	
	double alfa = 0.1;
	uint idx = 0;
	
	while(alfa > pow(10, -9)) {	
		curX = nxtX;
		curY = nxtY;
		double maxD = 0;
		
		for(uint i=0; i<N; i++) {
			double d = pow(X[i] - curX, 2) + pow(Y[i] - curY, 2);
			
			if(maxD < d) {
				maxD = d;
				idx = i;
			}
		}
		
		nxtX = (1 - alfa) * curX + alfa * X[idx];
		nxtY = (1 - alfa) * curY + alfa * Y[idx];
		alfa = alfa * 0.999999;
	
	}
	
	double ans = sqrt(pow(X[idx] - curX, 2) + pow(Y[idx] - curY, 2));
	cout << setprecision(10) << ans;
	
	return 0;
}
