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 (GCC 5.4.1) (judge verdict: Accepted)

// Ver19.03
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define INF (1 << 29) * 2
#define LLINF 4545454545454545454
#define MOD 1000000007
#define ll long long
#define ull unsigned long long
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
int upll(const void *a, const void *b) { return *(ll *)a < *(ll *)b ? -1 : *(ll *)a > *(ll *)b ? 1 : 0; }
int downll(const void *a, const void *b) { return *(ll *)a < *(ll *)b ? 1 : *(ll *)a > *(ll *)b ? -1 : 0; }
void sortup(ll *a, int n) { qsort(a, n, sizeof(ll), upll); }
void sortdown(ll *a, int n) { qsort(a, n, sizeof(ll), downll); }

int main()
{
  double n;
  scanf("%lf", &n);
  printf("%lf\n", ceil(n / 2) / n);
}