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 <algorithm>
#include <vector>
#include <string>
#include <numeric>
#include <map>
#include <iomanip>
#include <limits.h>

using namespace std;    
typedef long long ll;
typedef pair<ll, ll> pll;
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
const bool DEBAG = false;

int main(){

    int N;
    cin >> N;

    vector<double> v(N);
    for(int i = 0; i < N; i++) cin >> v[i];

    double mul = 1.0;
    for(int i = 0; i < N; i++){
        mul *= v[i];
    }

    double div = 0;
    for(int i = 0; i < N; i++){
        div += (mul / v[i]);
    }

    cout << setprecision(10) << mul / div << endl;

}