Problem:
Score : 100 points

Problem Statement

Given is an integer N.

Takahashi chooses an integer a from the positive integers not greater than N with equal probability.

Find the probability that a is odd.

Constraints

1 \leq N \leq 100

Input

Input is given from Standard Input in the following format:

N

Output

Print the probability that a is odd.
Your output will be considered correct when its absolute or relative error from the judge's output is at most 10^{-6}.

Sample Input 1

4

Sample Output 1

0.5000000000

There are four positive integers not greater than 4: 1, 2, 3, and 4. Among them, we have two odd numbers: 1 and 3. Thus, the answer is \frac{2}{4} = 0.5.

Sample Input 2

5

Sample Output 2

0.6000000000

Sample Input 3

1

Sample Output 3

1.0000000000

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long; 
using pii =  pair<int, int>;
using pll =  pair<long long, long long>;
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define all(x) (x).begin(),(x).end()
constexpr char ln =  '\n';
constexpr long long MOD = 1000000007LL;
//constexpr long long MOD = 998244353LL;
template<class T, class U> inline bool chmax(T &a, U b) { if (a < b) { a = b; return true;} return false; }
template<class T, class U> inline bool chmin(T &a, U b) { if (a > b) { a = b; return true;} return false; }
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
 
int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    
    int n; cin >> n;
    double ans = 0;
    
    if(n%2==0) ans = (double)n/2/n;
    else ans = (double)(n+1)/2/n;
    
    cout << ans << ln;
    return 0;
}