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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef string st;
typedef vector<int> vi;
typedef vector<st> vs;
#define rep(i, n) for(int i = 0; i < n; i++)
#define fogg(i,a,b) for(int i = (a); i <= (b); i++)
#define ford(i,a,b) for(int i = (a); i >= (b); i--)
#define test int t; cin >> t; while(t--)
#define debug(x) cout << '>' << #x << ':' << x << "\n";
#define endl '\n'
#define pb push_back
#define gp " "
#define ff first
#define ss second
#define mp make_pair



int main() {
    int n;
    cin >> n;
    int a[n];
    rep(i,n){
        cin >> a[i];
    }
    float sum=0;
    rep(i,n){
        sum+=1/a[i];
    }
    cout << 1/sum;

       
    

return 0;
}