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;
#define ll long long int
#define llu unsigned long long int
#define Endl endl
#define mod  998244353
#define mkp make_pair
#define pb push_back
#define cin(n) scanf("%d",&n)
#define cout(n) printf("%d ",n)
#define sf(n) scanf("%d",&n)
#define pf(n) printf("%d ",n)
#define bitget(n) __builtin_popcountll(n)
ll dx[] = {1,1,0,-1,-1,-1,0,1};
ll dy[] = {0,-1,-1,-1,0,1,1,1};
ll dxx[] = {1,-1,0,0};
ll dyy[] = {0,0,-1,1};
ll modexpo(ll x,ll y){if(y==0)return 1;if(y%2){ll viky=modexpo(x,y/2);
return (((x*viky)%mod)*viky)%mod;}else{ll viky=modexpo(x,y/2);return (viky*viky)%mod;
}}
// ll intpoint(ll x1, ll y1, ll x2, ll y2) { return ( __gcd(abs(x1-x2),abs(y1-y2))+1); }
// bool isvalid(ll x,ll y,ll n,ll m) { return (x>=0&&x<n&&y>=0&&y<m); }
// mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// int pp = uniform_int_distribution<int>(0, kk-1)(rng);
// ---------------------///////////////---------------------------------
// ---------------------///////////////----------------------------------
// -------------/////////////////---------------------------------///////



int main() {
    // #ifndef ONLINE_JUDGE
    //     freopen("input.txt", "r", stdin);
    //     freopen("output.txt", "w", stdout);
    // #endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    ll n;
    cin>>n;
    ll ap[n+11];
    for(ll i=0;i<n;i++){
        cin>>ap[i];
    }

    long double ans = 0;

    for(ll i=0;i<n;i++){
        ans += (long double)((long double)(1)/(long double)(ap[i]));
    }

    ans = (long double)(1)/(long double)(ans);

    // set
    cout<<setprecision(15)<<ans<<fixed;
    
    return 0;
}