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: Runtime Error)

#include<math.h>
#include<map>
#include<iostream>
#include <algorithm>
#include <string>
#include<vector>
#include<array>
#include<tuple>
#include<queue>
#include<stack>
#include<iomanip>
#include<bitset>
#include <list>
#include<numeric>
#include <fstream>
#include<set>
//AtCoderC++.cpp

using namespace std;

template<typename T>
inline T ceiling(T n, T d) {
	return n % d == 0 ? n / d : n / d + 1;
}


int main() {
	int N; cin >> N;
	auto as = vector<int>(N);
	long long mul = 1;
	for (size_t i = 0; i < N; i++) {
		cin >> as[i];
		mul *= as[i];
	}
	auto denom = 0;
	for (size_t i = 0; i < N; i++) {
		denom += mul / as[i];
	}
	cout << mul / denom << endl;
}