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>
int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
//    return *(long long*)a - *(long long*)b;
}
//  qsort(A, b, sizeof(int), compare_int);

long long  factorial(long long n) {
    if (n > 0) {
        return n * factorial(n - 1);
    } else {
        return 1;
    }
}

int min(int a, int b) {
    if (a > b)
        return b;
    else
        return a;
}

int distance(int a, int b) {
  return abs(a-b)*abs(a-b);
}

//int ()



int main()
{
  int a ,b, c, d, e, X, Y;
  double d_a,d_b, d_temp, d_count;
//  long long  a, b, c, d, e, X, Y;
  //  double  c;
//  int  A[200003]={};
  long long  A[100003]={};
  long long  B[100003]={};
  int  C[100003][11]={};

  int  flg=0;

//  long long  a, b, c, d, X, Y;
  char s[101][101]={};
  char t[101]={};
//  int N, M;
//  int A[1000000001]={};
//  long long  C[101]={};
  int i,j,k;
  long long p,q;
//  int flag=0;
//  int tmp=0;
//  char tmp[101]={};
//  long long min=10000000000;
//  long long min=100000001;
  int len=0;
  int l=0;
  int r=1000001;
  int cnt=0;
  int num=0;
  int num1=0;
  int num2=0;
  int ans=0;
  int max=0;
  int sum=0;
  int addr=0;
  int ptn=0;
  long long  tmp=0;
//  bool on_off[100]={};
  
//  char alp[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

  //  a=0;
//  scanf("%lld %lld", &a, &b);
  scanf("%d", &a);
  scanf("%lf", &d_a);
  d_count=1;
  for(i=1;i<a;i++){
    scanf("%lf", &d_b);
    d_count=d_count*d_b+d_a;
    d_a=d_a*d_b;
//    printf("d_a=%lf d_count=%lf\n",d_a,d_count);
  }

  
//  if(a<3200)  printf("red\n");
//  else   printf("%s\n",t);
  printf("%lf\n",d_a/d_count);
 

return 0;
}