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)

/*
	submission # - User: herp_sy
	https://atcoder.jp/contests/
*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <stdbool.h>

#define _CRT_SECURE_NO_WARNINGS
#define TLong long long
#define TBMod 1000000007

int collatzProblem(int n){
	return ((n % 2) ? (3 * n + 1) : (n / 2));
}

int main(int argc, char const *argv[])
{
	int cir[503];

	scanf("%d",&cir[1]);
	for (int i = 2; i < 500; ++i)	cir[i] = collatzProblem(cir[i - 1]);
	for (int i = 1; i < 500; ++i)
	{
		for (int j = i + 1; j < 500; ++j)
		{
			if(cir[i] == cir[j]){
				printf("%d\n",j);
				return 0;
			}
		}
	}
}