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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <ios>
#include <iostream>
#include <vector>

int gcd(int a, int b) {
  if (b == 0)
    return a;
  return gcd(b, a % b);
}

int lcm(int a, int b) { return (a / gcd(a, b)) * b; }

int main() {
  int n;
  std::cin >> n;

  std::vector<int> a(n);
  for (int i = 0; i < n; ++i)
    std::cin >> a[i];

  int m = 1;
  for (int i = 0; i < n; ++i)
    m = lcm(m, a[i]);

  int c = 0;
  for (int i = 0; i < n; ++i)
    c += m / a[i];

  std::cout << std::fixed << double(m) / c << std::endl;
}