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)

//
//  main.cpp
//  B - Resistors in Parallel
//
//  Created by Hashizo on 2019/08/18.
//  Copyright © 2019 Hashizo. All rights reserved.
//

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <regex>

using namespace std;

#undef LONG_MAX
#undef LONG_MIN

#define LONG_MAX	9223372036854775807
#define LONG_MIN	(-LONG_MAX  -1)

#define ll long long

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)

#define SORT(c) sort((c).begin(), (c).end())
#define REVERSE(v) reverse((v).begin(), (v).end())

int main(int argc, const char * argv[]) {
	double pulus = 0.0;
	ll n;
	cin >> n;

	double a;
	REP(i, n) {
		cin >> a;
		pulus = pulus + (1/a);
	}

	
	cout << 1/pulus << endl;
	return 0;
}