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)

// in the name of Allah

#include <bits/stdc++.h>
using namespace std;

#define endl '\n'
#define pb push_back
#define ll long long
#define double long double
#define sz(x) ((int) (x.size()))
#define all(x) (x).begin(), (x).end()

#define fr first
#define se second

const ll mod = 1000000007;

inline ll add(ll a, ll b) {
	a += b;
	if(a >= mod) return a - mod;
	return a;
}

inline ll mul(ll a, ll b) {
	a *= b;
	if(a >= mod) return a % mod;
	return a;
}

inline ll power(ll a, ll b) {
	ll p = 1;
	while(b) {
		if(b&1) p = mul(p, a);
		a = mul(a, a);
		b /= 2;
	}
	return p;
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);

	int n;
	cin >> n;
	ll a[n];
	for(int i=0; i<n; i++) cin >> a[i];
	double res = 0;
	for(int i=0; i<n; i++) {
		res += 1.0 / a[i];
	}
	cout << fixed << setprecision(12) << 1/res << endl;

	return 0;
}