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<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<tuple>
#include<iomanip>
#include<map>
#include<queue>
#define int long long
using namespace std;
const int INF = 1e10;
const int ZERO = 0;
int gcd(int a,int b) {
	if(a < b) {
		int tmp = a;
		a = b;
		b = tmp;
	}
	int r = a % b;
	while(r != 0) {
		a = b;
		b = r;
		r = a % b;
	}
	return b;
}
signed main() {
	int N;
	cin >> N;
	vector<int> vec(N);
	for(int i = 0;i < N;i++) {
		cin >> vec.at(i);
	}
	double L = vec.at(0);
	for(int i = 1;i < N;i++) {
		int tmp = gcd(L,vec.at(i));
		L = L * vec.at(i) / tmp;
	}
	double G = 0;
	for(int i = 0;i < N;i++) {
		G += L / vec.at(i);
	}
	double answer = L / G;
  cout << fixed; 
	cout << setprecision(9) << answer << endl;
}