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;
typedef long long ll;
#define rep(i, srt, end) for (long long i = (srt); i < (long long)(end); i++)

constexpr ll ITRMAX = 100;
ll n;
vector<double> x, y;

double r(double x0, double y0) {
    double res = 0;
    rep(i, 0, n){
        double dx = x[i] - x0;
        double dy = y[i] - y0;
        double dr = dx * dx + dy * dy;
        res = max(res, sqrt(dr));
    }
    return res;
}

double f(double x){
    double ly = -1000, ry = 1000;
    rep(itr, 0, ITRMAX){
        double my1 = (2 * ly + ry) / 3;
        double my2 = (ly + 2 * ry) / 3;
        if(r(x, my1) > r(x, my2)) ly = my1;
        else ry = my2;
    }
    return r(x, ly);
}

int main(){
    cin >> n;
    x.resize(n);
    y.resize(n);
    rep(i, 0, n) cin >> x[i] >> y[i];
    double lx = -1000, rx = 1000;
    rep(itr, 0, ITRMAX){
        double mx1 = 2 * lx / 3 + rx / 3;
        double mx2 = lx / 3 + 2 * rx / 3;
        if(f(mx1) > f(mx2)) lx = mx1; 
        else rx = mx2;
    }
    printf("%.20lf\n", f(lx));
    return 0;
}