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;

using ll = long long;

const int mod = 1000000007;
const int INF = 1000000000;
const double EPS = 1e-9;

struct Point {
    double x, y;
};

int N;
vector<Point> p;

double dist(Point a, Point 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 res = 0;
    for (int i = 0; i < N; ++i) {
        res = max(res, dist((Point){x, y}, p[i]));
    }
    return res;
}

double g(double x) {
    double l = 0, r = 1000;
    for (int i = 0; i < 100; ++i) {
        double c1 = (l*2 + r) / 3.0;
        double c2 = (l + r*2) / 3.0;
        if (f(x, c1) > f(x, c2)) l = c1;
        else r = c2;
    }
    return f(x, l);
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> N;
    p = vector<Point>(N);
    for (int i = 0; i < N; ++i) cin >> p[i].x >> p[i].y;
    double l = 0, r = 1000;
    for (int i = 0; i < 100; ++i) {
        double c1 = (l*2 + r) / 3.0;
        double c2 = (l + r*2) / 3.0;
        if (g(c1) > g(c2)) l = c1;
        else r = c2;
    }
    double ans = g(l);
    printf("%.10f\n", ans);
}