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: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"os"
)

var sc = bufio.NewScanner(os.Stdin)

const inf = 1 << 60
const mod int = 1e9 + 7

func ri() (n int) {
	sc.Scan()
	for _, v := range sc.Bytes() {
		n = n*10 + int(v-48)
	}
	return
}

func mul(a []int) int { // multple
	res := 1
	for _, v := range a {
		res *= v
	}
	return res
}

func main() {
	sc.Split(bufio.ScanWords)
	n := ri()

	a := make([]int, n)
	for i := 0; i < n; i++ {
		a[i] = ri()
	}

	if len(a) == 1 {
		fmt.Println(a[0])
		return
	}

	bunsi := float64(mul(a))

	bunbo := 0.0
	m := float64(mul(a))
	for _, aa := range a {
		bunbo += m / float64(aa)
	}

	ans := bunsi / bunbo

	fmt.Println(ans)
}