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

template <typename T> T load() { T r; cin >> r; return r; }
template <typename T> vector<T> loadMany(int n) { vector<T> rs(n); generate(rs.begin(), rs.end(), &load<T>); return rs; }

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	auto n = load<int>();
	auto xs = loadMany<int>(n);
	auto s = multiset<long double>();
	for (auto x : xs)
		s.insert(1.0l / x);
	while (s.size() > 1) {
		auto x1 = *s.begin();
		s.erase(s.begin());
		auto x2 = *s.begin();
		s.erase(s.begin());
		s.insert(x1 + x2);
	}
	auto answer = 1.0l / *s.begin();
	cout << fixed << setprecision(8) << answer << '\n';
}