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>
#include<string.h>
#include<stdlib.h>
#include<math.h>
long long cmpfunc (const void * a, const void * b) {
   return ( *(long long*)a - *(long long*)b );
}



int main(void){
   
long long int test,i,j,n,count,flag=0,o1=0,o2=0,b1,x,m,l,max,sum2,min,f,c,r,o,sum1,sum=0,y1,y2,a[1000000],b,count1=0,a2,b2;
float k;
scanf("%lld",&n);
scanf("%lld",&a[0]);
sum1=sum2=0;
y1=y2=1;
for(i=1;i<n;i++){
    scanf("%lld",&a[i]);
    if(a[i-1]==a[i]){
        y1+=1;
        
    }else{
        y1*=a[i];
        y1+=a[i-1];
        a[i]*=a[i-1];
    }
   // printf("%lld %lld\n",a[i],y1);
}
k=((a[n-1]*(1.0))/(y1*1.0))*(1.0);

printf("%.50f",k);
return 0;
}