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 <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

int N,x[301],y[301];

double get_r(double X, double Y){
    double ret = 0;

    for(int i = 0;i < N;++i)
        ret = max(ret,(X - x[i]) * (X - x[i]) + (Y - y[i]) * (Y - y[i]));

    return ret;
}

double searchY(double X){
    double lo = 0,hi = 1000;

    for(int it = 0;it < 1000;++it){
        double a = (2 * lo + hi) / 3;
        double b = (lo + 2 * hi) / 3;

        double ra = get_r(X,a);
        double rb = get_r(X,b);

        if(ra < rb) hi = b;
        else lo = a;
    }

    return get_r(X,lo);
}

double searchX(){
    double lo = 0,hi = 1000;

    for(int it = 0;it < 1000;++it){
        double a = (2 * lo + hi) / 3;
        double b = (lo + 2 * hi) / 3;

        double ra = searchY(a);
        double rb = searchY(b);

        if(ra < rb) hi = b;
        else lo = a;
    }

    return searchY(lo);
}

int main(){
    scanf("%d",&N);

    for(int i = 0;i < N;++i)
        scanf("%d %d",&x[i],&y[i]);

    printf("%.7f\n",sqrt(searchX()));

    return 0;
}