Problem:
Score : 200 points

Problem Statement

Given is a sequence of N integers A_1, \ldots, A_N.

Find the (multiplicative) inverse of the sum of the inverses of these numbers, \frac{1}{\frac{1}{A_1} + \ldots + \frac{1}{A_N}}.

Constraints

1 \leq N \leq 100

1 \leq A_i \leq 1000

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \ldots A_N

Output

Print a decimal number (or an integer) representing the value of \frac{1}{\frac{1}{A_1} + \ldots + \frac{1}{A_N}}.

Your output will be judged correct when its absolute or relative error from the judge's output is at most 10^{-5}.

Sample Input 1

2
10 30

Sample Output 1

7.5

\frac{1}{\frac{1}{10} + \frac{1}{30}} = \frac{1}{\frac{4}{30}} = \frac{30}{4} = 7.5.

Printing 7.50001, 7.49999, and so on will also be accepted.

Sample Input 2

3
200 200 200

Sample Output 2

66.66666666666667

\frac{1}{\frac{1}{200} + \frac{1}{200} + \frac{1}{200}} = \frac{1}{\frac{3}{200}} = \frac{200}{3} = 66.6666....

Printing 6.66666e+1 and so on will also be accepted.

Sample Input 3

1
1000

Sample Output 3

1000

\frac{1}{\frac{1}{1000}} = 1000.

Printing +1000.0 and so on will also be accepted.

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
 
public class Coder
{

    public static int[] ReadArray()
    {
        return Array.ConvertAll(Console.ReadLine().Split(' '), m => Convert.ToInt32(m));
    }

    public static int ReadInt()
    {
        return Convert.ToInt32(Console.ReadLine());
    }

    public static string ReadStr()
    {
        return Console.ReadLine();
    }

    public static long ReadLong()
    {
        return Convert.ToInt64(Console.ReadLine());
    }

    public static long[] ReadLongArray()
    {
        return Array.ConvertAll(Console.ReadLine().Split(' '), m => Convert.ToInt64(m));

    }

    public static void Main()
    {

        var a = ReadInt();
        int[] A = ReadArray();

        double sum = 0;
        foreach(var i in A)
        {
            sum += (double)1 / i;
        }

        Console.WriteLine((double)1 / sum);

    }

}