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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main (args : Array<String>) {
	val n = readLine()
	val N : Double = n!!.toDouble()
	var count : Double = 0.0
	var result : Double = 0.0
	for (i in 1..N.toInt()) {
		if (i % 2 == 1) {
			count += 1.0
		}
	}
	result = count / N
	println("${result}")
}