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<math.h>
#define loop(i, l, r) for(i=l; i<r; i++)/*Instead of function, (for).*/
#define rep(i,r) for(i=0; i<r; i++)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long int lli;
/*Return Combination ,(nCr)*/
int combi(int n, int r)
{
	if(r==0 || r==n) return 1;
	else return combi(n-1,r) + combi(n-1,r-1);
}
/*Return gcd(a, b)*/
int gcd(int a, int b){
	if(b == 0) return a;
	else return gcd(b, a % b);
}
int main(void)
{
	int n, i;
	scanf("%d", &n);
	double a[n], k=0.0, ans;

	rep(i,n){
		scanf("%lf", &a[i]);
		k += 1.0/a[i];
	}
	ans = 1.0/k;
	printf("%.10f", ans);
	return 0;
}