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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <iomanip>
#include <fstream>
#include <vector>
#include <algorithm>
#include <math.h>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>

#define _GLIBCXX_DEBUG
using namespace std;
using vi = vector<int>;
using vvi = vector<vi>;

int main(int argc, const char * argv[]) {
    // fasten cin
    cin.tie(0);
    ios::sync_with_stdio(false);
    // implement
    int64_t n;
    cin >> n;
    vi a(n);
    for(int i=0; i<n; i++){
        cin >> a.at(i);
    }
    double tmp = 0;
    for(int i=0; i<n; i++){
        tmp += 1.0/a.at(i);
    }
    cout << fixed << setprecision(6);
    cout << 1/tmp << endl;
}