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)

// SeeAlso: https://atcoder.jp/contests/abc138/tasks/abc138_b

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define MAX 100

typedef int _loop_int;
#define REP(i,n) for(int i = 0; i < n; i++)
#define FOR(i,a,b) for(_loop_int i=(_loop_int)(a);i<(_loop_int)(b);++i)
#define FORR(i,a,b) for(_loop_int i=(_loop_int)(b)-1;i>=(_loop_int)(a);--i)

#define DEBUG(x) cout<<#x<<": "<<x<<endl
#define DEBUG_VEC(v) cout<<#v<<":";REP(i,v.size())cout<<" "<<v[i];cout<<endl
#define ALL(a) (a).begin(),(a).end()

#define CHMIN(a,b) a=min((a),(b))
#define CHMAX(a,b) a=max((a),(b))

// 最大公約数
inline constexpr ll gcd(ll a,ll b){if(!a||!b)return 0;while(b){ll c=b;b=a%b;a=c;}return a;}
// 最小公倍数
inline constexpr ll lcm(ll a,ll b){if(!a||!b)return 0;return a*b/gcd(a,b);}

#define print2D(h, w, arr) REP(i, h) { REP(j, w) cout << arr[i][j] << " "; cout << endl; }
template<class T> void print(const T& x){cout << setprecision(12) << x << endl;}
template<class T, class... A> void print(const T& first, const A&... rest) { cout << first << " "; print(rest...); }

int N;

int main() {
    cin >> N;
    double s = 0;

    REP(i, N) {
        int a;
        cin >> a;
        double temp = 1.0/a;
        s += temp;
    }

    double ans = 1.0 / s;
    print(ans);
    return 0;
}