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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
#define DUMP(x)  cout << #x << " = " << (x) << endl;
#define FOR(i, m, n) for(ll i = m; i < n; i++)
#define IFOR(i, m, n) for(ll i = n - 1; i >= m; i-- )
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
#define FOREACH(x,a) for(auto& (x) : (a) )
#define ALL(v) (v).begin(), (v).end()

int main(){
  ll n; cin >> n;
  vector<ll> a(n);
  REP(i,n) cin >> a[i];
  double ans = 1.0;
  double bunbo = 0.0;
  REP(i,n) bunbo += 1.0/a[i];
  cout << fixed << setprecision(10);
  cout << 1.0 / bunbo << endl;
}