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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#define max(a, b)	(((a) > (b)) ? (a) : (b))		/* ２個の値の最大値 */
#define min(a, b)	(((a) < (b)) ? (a) : (b))		/* ２個の値の最小値 */
#define ENTER		printf("\n")					/* 改行プリント */
int DBG = 1;										/* デバッグプリント 提出時は0 */
/* main *************************************************************/
int main()
{
	int		i,j,k,n,m,x,y;
	int		a[256];
	double	ans = 0;
	
	scanf("%d", &n);
	for(i=0;i<n;i++) {
		scanf("%d",&j);
		ans += (double)1/j;
	}
	
	printf("%.14lf\n", (double)1/ans);

	return 0;
}