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)

#include<iostream>
#include<vector>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<map>
#include<bitset>
using namespace std;

typedef long long ll;
#define rep(i, n) for(int i = 0;i < (n);i++)
#define repr(i, n) for(int i = (n);i >= 0;i--)
#define repf(i, m, n) for(int i = (m);i < (n);i++)
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1;} return 0;}
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1;} return 0;}
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
//////////////////////////////////////////////////


int main() {
    int n; cin >> n;
    double ans = (double)(n-n/2)/n;
    printf("%.6lf\n", ans);
    return 0;
}