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>
#define rep(i,n) for (long long i = 0; i < (n); ++i)
//g++ -std=c++11 

using namespace std;

long long N;
double X[55];
double Y[55];


double getget(double x, double y) {
	double ans = 0;
	rep(i, N) {
		ans = max(ans, sqrt((X[i] - x) * (X[i] - x) + (Y[i] - y) * (Y[i] - y)));
	}
	return ans;
}



// xを固定した場合の半径の最小
double cal(double x) {
	double starty = 0;
	double endy = 1000;

	double ans = 10001000;

	while(starty + 0.0000001 < endy) {
		double y1 = starty + (endy-starty)/3.0 * 1;
		double y2 = starty + (endy-starty)/3.0 * 2;

		double num1 = getget(x, y1);
		double num2 = getget(x, y2);

		if(num1 < num2) {
			endy = y2;
		} else {
			starty = y1;
		}
	}
	return getget(x, starty);
}

int main(){
	cin >> N;
	rep(i, N) cin >> X[i] >> Y[i];


	double startx = 0;
	double endx = 1000;

	double ans = 10001000;

	while(startx + 0.0000001 < endx) {
		double x1 = startx + (endx-startx)/3.0 * 1;
		double x2 = startx + (endx-startx)/3.0 * 2;

		double num1 = cal(x1);
		double num2 = cal(x2);
		ans = min(ans, num1);
		ans = min(ans, num2);

		if(num1 < num2) {
			endx = x2;
		} else {
			startx = x1;
		}
	}


	printf("%.20f\n", ans);

}
