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;
int main() {
	int n;
	cin >> n;
	vector<int> x;
	vector<int> y;
	x.resize(n);
	y.resize(n);
	double X = 0;
	double Y = 0;
	for(int i = 0; i < n; ++i) {
		cin >> x[i] >> y[i];
		X += x[i];
		Y += y[i];
	}
	X /= n;
	Y /= n;
	auto Rand = [&] () {
		return rand() % 10000 / 10000.0;
	};
	auto calc = [&] (double X, double Y) {
		double ret = 0;
		for(int i = 0; i < n; ++i) {
			ret = max(ret, sqrt((X - x[i]) * (X - x[i]) + (Y - y[i]) * (Y - y[i])));
		}
		return ret;
	};
	double T = 10000.0;
	while(T > 0.0000001) {
		double new_X = X + T * (Rand() * 2 - 1.0);
		double new_Y = Y + T * (Rand() * 2 - 1.0);	
		if(calc(X, Y) > calc(new_X, new_Y)) {
			X = new_X;
			Y = new_Y;
		}
		T *= 0.999991;
	}
	cout << fixed << setprecision(15);
	cout << calc(X, Y) << '\n';
	return 0;
}