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 (int i = 0; i < (n); ++i)
#define sz(x) int(x.size())
#define show(x) {for(auto i: x){cout << i << " ";} cout << endl;}
using namespace std;
using ll = long long;
using P = pair<int, int>;

int N;
struct crd {double x, y;};
vector<crd> dots;

double dist(crd a, crd b) {
    double dx = a.x - b.x;
    double dy = a.y - b.y;
    return sqrt(dx * dx + dy * dy);
}

double f(double x, double y) {
    double d = 0;
    rep(i, N) {
        d = max(d, dist((crd){x, y}, dots[i]));
    }
    return d;
}

double g(double x) {
    double left = 0, right = 1000;
    rep(i, 80) {
        double l2 = (left * 2 + right) / 3;
        double r2 = (left + right * 2) / 3;
        if (f(x, l2) < f(x, r2)) right = r2;
        else left = l2;
    }
    return f(x, left);
}

int main() {
    cin >> N;
    dots = vector<crd> (N);
    rep(i, N) cin >> dots[i].x >> dots[i].y;
    double left = 0, right = 1000;
    rep(i, 80) {
        double l2 = (left * 2 + right) / 3;
        double r2 = (left + right * 2) / 3;
        if (g(l2) < g(r2)) right = r2;
        else left = l2;
    }
    double ans = g(left);
    printf("%.10f\n", ans);
    return 0;
    // ll, 0, -, 1i, for s&g, debug
}