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 Go (1.6) (judge verdict: Accepted)

// 通分して
// package main

// import "fmt"

// func main() {
// 	var n int
// 	fmt.Scan(&n)
// 	a := make([]float64, n)
// 	for i := 0; i < n; i++ {
// 		fmt.Scan(&a[i])
// 	}

// 	x := 1.0 // 分母
// 	for _, aa := range a {
// 		x *= aa
// 	}

// 	y := 0.0 // 分子
// 	for _, aa := range a {
// 		y += x / aa
// 	}

// 	ans := x / y
// 	fmt.Println(ans)
// }

// ------------------------
// ナイーブに
package main

import "fmt"

func main() {
	var n int
	fmt.Scan(&n)
	a := make([]float64, n)
	for i := 0; i < n; i++ {
		fmt.Scan(&a[i])
	}

	x := 0.0 // 分母
	for _, aa := range a {
		x += 1 / aa
	}

	ans := 1 / x
	fmt.Println(ans)
}