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)

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

#define all(a) a.begin(), a.end()
#define ff first
#define ss second
#define pb push_back

using ll = long long;
using pii = pair<int, int>;

template<typename T> using V = vector<T>;

int main() {
	ios::sync_with_stdio(false);cin.tie(nullptr);
	int n; cin >> n;
	vector<ll> a(n);
	ll prod = 1;
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
		prod *= a[i];
	}
	vector<ll> left(n);
	ll sum = 0;
	for (int i = 0; i < n; ++i) {
		left[i] = prod / a[i];
		sum += left[i];
	}
	cout << fixed << setprecision(10);
	cout << (long double)prod / sum;
}