Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

2
0 0
1 0

Sample Output 1

0.500000000000000000

Both points are contained in the circle centered at (0.5,0) with a radius of 0.5.

Sample Input 2

3
0 0
0 1
1 0

Sample Output 2

0.707106781186497524

Sample Input 3

10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9

Sample Output 3

6.726812023536805158

If the absolute or relative error from our answer is at most 10^{-6}, the output will be considered correct.

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;

class Program
{
    static void Main()
    {
        string[] arr = Console.ReadLine().Split(' ');
        int N = int.Parse(arr[0]);
        decimal max = decimal.MinValue;
        int[][] xy = new int[N][];

        for(int i = 0; i < N; i++)
        {
            string[] a = Console.ReadLine().Split(' ');
            xy[i] = new int[]{ int.Parse(a[0]), int.Parse(a[1]) };
        }

        for(int i = 0; i < N; i++)
        {
            for(int j = i + 1; j < N; j++)
            {
                max = Math.Max(max, Distance(xy[i], xy[j]));
            }
        }
        Console.WriteLine(max / 2m);
    }

    private static decimal Distance(int[] pointA, int[] pointB)
    {
        return Convert.ToDecimal(Math.Sqrt((double)(Convert.ToDecimal(Math.Pow(pointA[0] - pointB[0], 2)) + Convert.ToDecimal(Math.Pow(pointA[1] - pointB[1], 2)))));
    }
}