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>
using namespace std;
#define all(V) V.begin(),V.end()
#define ll long long
const ll MOD = 1000000007;

complex<double> P[51], Q[51];
int N;

double F(double X) {
	complex<double> A = { X,0 };
	double mf = 0;
	for (int i = 0;i < N;i++) {
		mf = max(mf, abs(Q[i] - A));
	}
	return mf;
}

int main() {
	cin >> N;
	double x, y;
	for (int i = 0;i < N;i++) {
		cin >> x >> y;
		P[i] = { x,y };
	}
	double big, small, id, jd, l, r, l3, r3;
	complex<double> M, me;

	vector<double> ans;

	for (int i = 0;i < N;i++) {
		for (int j = i + 1;j < N;j++) {
			M = (P[i] + P[j]) / 2.0;
			me = complex<double>(0, 1) * (P[i] - P[j]) / abs(P[i] - P[j]);
			for (int k = 0;k < N;k++) {
				Q[k] = (P[k] - M) / me;
			}
			l = -4000, r = 4000;
			
			for (int a = 0;a < 100;a++) {
				l3 = (l * 2 + r) / 3;
				r3 = (l + r * 2) / 3;
				if (F(l3) > F(r3)) {
					l = l3;
				}
				else {
					r = r3;
				}
			}

			ans.push_back(F(l3));
		}
	}
	sort(all(ans));
	printf("%.10f\n", ans[0]);
}