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>
         
        int main(){
          	int a;
          	int n[200000];
          	int *n_pointer = n;
          
          
        	char buf[200];
          	char buf2[300000];
          	fgets(buf, 20000, stdin);
          	fgets(buf2, 200000, stdin);
          	sscanf(buf,"%d", &a);
          
          	char *tp;
          	tp = strtok(buf2, " ");
          	int temporaryint = *tp - '0';
          	
          	n[0] = temporaryint;
          	int i = 1;
          	while(tp = strtok(NULL, " ")){
              	int temporaryint = *tp - '0';
              	n[i] = temporaryint;
              	i++;
            }
          	
          	int n_length = 0;
          	for(int i = 0; n[i] != 0; i++){
            	n_length++;
            }
            
          	float sum = 0;
          	for(int i = 0; i < n_length; i++){
            	sum = sum + 1/(float)n[i];
            }
          	printf("%f",1/sum);
          
          
        }