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;
int N;
vector<int> x, y;
double const eps = 1e-10;

double g(double X, double Y) {
    double ret = 0;
    for(int i=0; i<N; ++i) {
        ret = max(ret, sqrt((x[i] - X) * (x[i] - X) + (y[i] - Y) * (y[i] - Y)));
    }
    return ret;
}

double f(double X) {
    double yl = 0, yr = 1000;
    while(yr - yl > eps) {
        double left = (2 * yl + yr) / 3;
        double right = (2 * yr + yl) / 3;
        if(g(X, left) > g(X, right) + eps) yl = left;
        else yr = right;
    }
    return yl;
}

double solve() {
    double xl = 0, xr = 1000;
    while(xr - xl > eps) {
        double left = (2 * xl + xr) / 3;
        double right = (2 * xr + xl) / 3;
        if(g(left, f(left)) > g(right, f(right)) + eps) xl = left;
        else xr = right;
    }
    return g(xl, f(xl));
}

int main() {
    cin >> N;
    x.resize(N); y.resize(N);
    for(int i=0; i<N; ++i) cin >> x[i] >> y[i];
    cout << fixed << setprecision(10) << solve() << endl;
}