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 <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <iomanip>
using namespace std;
using ll = long long;
#define rep(i, j, n) for(int i = (j); i < (n); ++i)
#define rrep(i, j, n) for(int i = (n) - 1; (j) <= i; --i)
constexpr ll MOD = 1000000007;
constexpr ll INF = 1LL << 60;

int main() {
	cin.tie(0); ios_base::sync_with_stdio(false);

	int n; cin >> n;
	vector<double> x(n), y(n);
	double X = 0, Y = 0;
	rep(i, 0, n) {
		cin >> x[i] >> y[i];
		X += x[i];
		Y += y[i];
	}
	X /= (double)n;
	Y /= (double)n;

	auto dist = [](double x, double y) {
		return x * x + y * y;
	};

	double p = .1;
	double max_dist;
	rep(_, 0, 20000) {
		max_dist = 0.0;
		int max_idx = 0;
		rep(i, 0, n) {
			double d = dist(X - x[i], Y - y[i]);
			if (d > max_dist) {
				max_dist = d;
				max_idx = i;
			}
		}

		X += (x[max_idx] - X) * p;
		Y += (y[max_idx] - Y) * p;
		p *= 0.999;
	}
	cout << fixed << setprecision(10) << sqrt(max_dist) << endl;
	return 0;
}
