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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <limits.h>
#include <string.h>
#define ll long long int



int main()
{
   int s;
   scanf("%d",&s);
   int array[1000];
   memset(array,0,sizeof(array[0])*1000);
   array[1]=s;
    for (int i = 2;i<1000;i++)
    {
       if (array[i-1]%2==0)
       {
           array[i]=array[i-1]/2;
       }
       else if (array[i-1]%2!=0)
       {
          array[i]=3*array[i-1]+1;

       }

    }

   for (int j=1;j<1000;j++)
    for (int i = j+1;i<1000;i++)
    {
       if (array[j]==array[i])
       {
          printf("%d",i);
          return 0;

       }
    }





   return 0;
}