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 <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <stdlib.h>
#include <string.h>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <stdio.h>
#include <tuple>
#include <bitset>
#include <cfloat>
#include <fstream>
#include <limits.h>
#include <list>
#include <math.h>
#include <random>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <numeric>
#include <utility>
#include <sstream>
// *テンプレート*
#define int long long
#define rep(i, n) for (int i = 0; i < (n); i++)
#define P pair<int, int>
// aよりbの方が小さかったら、aの値をbの値に置き換える。(swapではない！)
template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
// aよりbの方が大きかったら、aの値をbの値に置き換える。
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}

using namespace std;
typedef long long ll;
map<int, int> mp;
using Graph = vector<vector<int>>; 
//最大公約数
ll gcd(ll a, ll b) {
	if (a < b)swap(a, b);
	if (b == 0) return a;
	return gcd(b, a % b);
}
//最小公倍数
ll lcm(ll a, ll b) {
	ll g = gcd(a, b);
	return a / g * b;
}

const long long MOD = 1000000007;
const long long INF = 1LL << 60;
// *テンプレートここまで*

// pair -> vector<pair<int,int>> 
//		   AandB.push_back(make_pair(a[i],b[i]))

// ^= -> ビットごとの排他的論理和を計算する

signed main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n; cin >> n;
	vector<double>a(n); rep(i, n)cin >> a[i];
	double sum = 0;
	rep(i, n) {
		sum += (1 / a[i]);
	}
	cout << fixed << setprecision(15) << 1 /sum << endl;
	return 0;
}