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: Wrong Answer)

#include <stdio.h>
typedef long long ll;

int gcd(ll a,ll b)
{
    int r=a%b;
    while(r!=0){
        a = b;
        b = r;
        r = a % b;
    }
    return b;
}

int main(){
    int n;
    ll x=1,g,sum=0;
    scanf("%d",&n);
    
    ll a[n];
    scanf("%lld",&a[0]);
    x=a[0];
    g=a[0];
    for(int i=1;i<n;i++){
        scanf("%lld",&a[i]);
        x*=a[i];
        g=(g>a[i])?gcd(g,a[i]):gcd(a[i],g);
    }
    if(n==1)g=1;
    ll lcm=x/g;
    for(int i=0;i<n;i++){
        sum+=lcm/a[i];
    }
    printf("%f",(double)lcm/sum);
}