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 <vector>
#include <array>
#include <algorithm>
#include <iostream>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <map>
#include <math.h>
#include <assert.h>
#include <queue>
#include <deque>
#include <iomanip>
#include <numeric>
#include <fstream>

using namespace std;

typedef long long ll;
typedef long double ld;

#define fi first
#define se second

#define mp make_pair
#define pb push_back
#define sz(x) ((int)(x).size())

typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<ld> vld;
typedef vector<vector<int>> vvi;
typedef vector<ll> vl;
typedef vector<vector<ll>> vvl;
typedef vector<vector<ld>> vvd;
typedef vector<string> vs;
typedef vector<bool> vb;

typedef pair <int, int> pii;
typedef pair <ll, ll> pll;



int main() {
    cout << setprecision(30);
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);

    ll n;
    cin >> n;
    vl a(n);
    ld kek = 0;
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        kek += 1. / (ld)(a[i]);
    }
    cout << (1 / kek) << endl;

    return 0;
}