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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php

$n = trim(fgets(STDIN));
$a = explode(' ', trim(fgets(STDIN)));

$sum = 0;
for($i = 0;$i < count($a);$i++) {
	$sum += 1 / (double)$a[$i];
}
echo 1 / $sum."\n";
// echo $tmp;
// $a = [];
// $sum = 0;
// for($i = 0;$i < $n;$i++) {
// 	$tmp = trim(fgets(STDIN));
// 	echo $tmp."\n";
// 	// $sum += (1 / $tmp);
// }

// echo sprintf('%.3f', 1/$sum);

// $arr = [$input[0]];
// $tmp = '';
// for($i = 1;$i < strlen($input);$i++) {
//   $m = count($arr);
//   $tmp .= $input[$i];

//   if($arr[$m - 1] === $tmp) {

//   } else {
//     $arr[] = $tmp;
//     $tmp = '';
//   }
// }

// echo count($arr)."\n";