Problem:
Score : 200 points

Problem Statement

A sequence a=\{a_1,a_2,a_3,......\} is determined as follows:

The first term s is given as input.

Let f(n) be the following function: f(n) = n/2 if n is even, and f(n) = 3n+1 if n is odd.

a_i = s when i = 1, and a_i = f(a_{i-1}) when i > 1.

Find the minimum integer m that satisfies the following condition:

There exists an integer n such that a_m = a_n (m > n).

Constraints

1 \leq s \leq 100

All values in input are integers.

It is guaranteed that all elements in a and the minimum m that satisfies the condition are at most 1000000.

Input

Input is given from Standard Input in the following format:

s

Output

Print the minimum integer m that satisfies the condition.

Sample Input 1

8

Sample Output 1

5

a=\{8,4,2,1,4,2,1,4,2,1,......\}. As a_5=a_2, the answer is 5.

Sample Input 2

7

Sample Output 2

18

a=\{7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1,4,2,1,......\}.

Sample Input 3

54

Sample Output 3

114

Solution in Ruby (2.7.1) (judge verdict: Accepted)


#1に収束すると予想されている関数で、さらに直前の数にしか依存していないので、循環は(4 2 1)を繰り返す。よってsが4,2,1以外なら被る数字は4
#4,2,1の場合は答えは4
def collatz(n)
  if n % 2 == 0
    return n/2
  else
    return 3 * n + 1
  end
end
s = gets.to_i
if s == 1 || s == 2 || s == 4
  puts 4
else
  n = collatz(s)
  arr = [n] #1個目の数を入れる
  four_flag = 0 #循環してない
  while true
    if n == 4 && four_flag == 0
      four_flag = 1 #はじめて4になったらフラグを立てる
    elsif n == 4
      break
    end
    n = collatz(n)
    arr.push(n)
  end
  puts arr.size + 1
end