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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
public class Main {	
	public static void main(String args[]){
		Scanner sc = new Scanner(System.in);
		int aa = sc.nextInt();
		double x[] = new double[aa];
		for(int n = 0; n < aa; n++){
			x[n] = sc.nextInt();
		}
		double a = 1;
		double b = x[0];
		for(int n=1; n < aa; n++){
			a = a*x[n] + b;
			b *= x[n];
			double t = f(Math.max(a, b), Math.min(a, b));
			a = a / t;
			b = b / t;
		}
		System.out.println((double)b/(double)a);
		
		
	}
	
	public static double f(double d, double e){
		double t = d % e;
		if(t == 1)return 1;
		else if(t == 0)return e;
		else return f(e, t);
	}
}