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 Rust (1.42.0) (judge verdict: Wrong Answer)

use std::io::Read;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();
    let a: Vec<usize> = (0..n).map(|_| iterator.next().unwrap().parse().unwrap()).collect();

    let mut l: usize = 1;
    for i in 0..n {
        l = lcm(l, a[i]);
    }

    let mut s: usize = 0;
    for i in 0..n {
        s += l / a[i];
    }

    let ans: f64 = l as f64 / s as f64;

    return format!("{}", ans);
}

/// 最大公約数（greatest common divisor）を求める
///
/// ユークリッドの互除法
fn gcd(mut x: usize, mut y: usize) -> usize {
    while y > 0 {
        let r = x % y;
        x = y;
        y = r;
    }
    return x;
}

/// 最小公倍数（least common multiple）を求める
///
/// m * n == gcd(m, n) * lcm(m, n)から、
/// lcm == m * n / gcd(m, n)
/// オーバーフロー対策として、除算を先に行う
fn lcm(m: usize, n: usize) -> usize {
    m / gcd(m, n) * n
}