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;
inline void rd(int &x){
  int k;
  int m=0;
  x=0;
  for(;;){
    k = getchar_unlocked();
    if(k=='-'){
      m=1;
      break;
    }
    if('0'<=k&&k<='9'){
      x=k-'0';
      break;
    }
  }
  for(;;){
    k = getchar_unlocked();
    if(k<'0'||k>'9'){
      break;
    }
    x=x*10+k-'0';
  }
  if(m){
    x=-x;
  }
}
inline void wt_L(char a){
  putchar_unlocked(a);
}
inline void wt_L(double x){
  printf("%.15f",x);
}
template<class S, class T> inline S divup_L(S a, T b){
  return (a+b-1)/b;
}
int main(){
  int N;
  double res;
  rd(N);
  res = (divup_L(N,2)) / (double)N;
  wt_L(res);
  wt_L('\n');
  return 0;
}
// cLay varsion 20190929-1

// --- original code ---
// {
//   int N;
//   double res;
//   rd(N);
//   res = (N /+ 2) / (double)N;
//   wt(res);
// }