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 Kotlin (1.0.0) (judge verdict: Time Limit Exceeded)

import java.util.*

fun main(args: Array<String>) {
    val sc = Scanner(System.`in`)
    val n = sc.nextLong()
    val a = (0 until n).map { sc.next().toLong() }
    println(problem138b(n, a))
}

fun problem138b(n: Long, a: List<Long>): Double {
    val max = a.max()
    val list = mutableListOf<Int>()
    a.forEach {
        var bai = 1
        while (true) {
            if (it * bai == max) {
                list.add(bai)
                return@forEach
            }
            bai++
        }
    }
    val sum = list.sum()
    return max!!.toDouble() / sum.toDouble()
}