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 <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#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;

double dist(double a, double b) {
	return a * a + b * b;
}

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

	int n; cin >> n;
	vector<double> x(n), y(n);
	double X = 0.0, Y = 0.0;
	rep(i, 0, n) {
		cin >> x[i] >> y[i];
		X += x[i];
		Y += y[i];
	}

	X /= (double)n; Y /= (double)n;
	double P = 0.1;
	double d, e;
	for (int i = 0; i < 30000; i++) {
		int f = 0;
		d = dist(X - x[0], Y - y[0]);
		for (int j = 1; j < n; j++) {
			e = dist(X - x[j], Y - y[j]);
			if (d < e) { d = e; f = j; }
		}
		X += (x[f] - X) * P;
		Y += (y[f] - Y) * P;
		P *= 0.999;
	}

	cout << fixed << setprecision(10) << sqrt(d) << endl;
	return 0;
}