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;
int x[58], y[58];

double sq(double d) {
    return d * d;
}

double d(double X, double Y) {
    double ret = 0.0;
    for (int i = 0; i < n; ++i) {
        ret = max(ret, sqrt(sq(X - x[i]) + sq(Y - y[i])));
    }
    return ret;
}

double cal(double X) {
    double lo = 0, hi = 1000;
    for (int i = 0; i < 100; ++i) {
        double m1 = (lo + lo + hi) / 3, m2 = (lo + hi + hi) / 3;
        if (d (X,m1) < d(X, m2)) hi = m2;
        else lo = m1;
    }
    return d(X, lo);
}

int main() {
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> x[i] >> y[i];
    }
    double lo = 0, hi = 1000;
    for (int i = 0; i < 100; ++i) {
        double m1 = (lo + lo + hi) / 3, m2 = (lo + hi + hi) / 3;
        if (cal(m1) < cal(m2)) hi = m2;
        else lo = m1;
    }
    printf("%.12f\n", cal(lo));
    return 0;
}