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)

/*
    Code written by Talant I.D.
*/
   
#include <bits/stdc++.h>
   
using namespace std;
   
typedef long long ll;
typedef pair <int, int> pii;
typedef tuple <int, int, int> tiii;
   
#define precision(n) fixed << setprecision(n)
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define mod (int)1e9+7
#define eps (double)1e-9
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define do_not_disturb ios::sync_with_stdio(0);cin.tie(0);
#define curMod 998244353

int main(){
		//do_not_disturb
		int n, i;
		cin >> n;
		double ans = 0;
		int a[n];
		
		for(i = 0; i < n; i++){
			cin >> a[i];
			ans += double(1./a[i]);
		}
		cout << double(1./ans);
		
    return 0;
} // What is to be written here?