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>
using namespace std;
typedef long long int ll;

// template {{{  0 
// using {{{ 1
using ll = long long int;
using pii = pair<int,int>;
using pll = pair<ll,ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vii = vector<pii>;
using vll = vector<pll>;
// }}} 1
// definition {{{ 1
// scaning {{{ 2
#define Scd(x) scanf("%d", &x)
#define Scd2(x,y) scanf("%d%d", &x, &y)
#define Scd3(x,y,z) scanf("%d%d%d", &x, &y, &z)
#define Scll(x) scanf("%lld", &x)
#define Scll2(x,y) scanf("%lld%lld", &x, &y)
#define Scll3(x,y,z) scanf("%lld%lld%lld", &x, &y, &z)
#define Scc(c) scanf("%c", &c);
#define Scs(s) scanf("%s", s);
#define Scstr(s) scanf("%s", &s);
// }}} 2
// constants {{{ 2
#define EPS (1e-7)
#define INF (2e9)
#define PI (acos(-1))
// }}} 2
// systems {{{ 2
#define Repe(x,y,z) for(ll x = z; x < y; x++)
#define Rep(x,y) Repe(x,y,0)
#define RRepe(x,y,z) for(ll x = y-z-1; x >= 0; x--)
#define RRep(x,y) RRepe(x,y,0)
// }}} 2
// output {{{ 2
#define YesNo(a) (a)?printf("Yes\n"):printf("No\n")
#define YESNO(a) (a)?printf("YES\n"):printf("NO\n")
// }}} 2
// }}} 1
// input {{{ 1
// }}} 1
// }}} 0

int main() {

    int N;
    Scd(N);

    int a;

    double ans = 0;

    Rep(i,N){
        Scd(a);
        ans += 1.0 / a;
    }

    printf ("%lf\n", 1/ans);
    return 0;
}