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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <algorithm>

#define REP(i, n) for (long long i = 0; i < (n); i++) 
#define ll long long

using namespace std;
//        tmp = stoi(S.substr(i,3));
int main(){
   
    ll l,s;

    vector <ll> ima;
    bool flg = false;

    cin >> s;

    l = s;
    ima.push_back(l);
    ll cnt = 1;
    while(flg == false)
    {

        if(l % 2 == 0) l = l/2;
        else l = 3 * l + 1;

        for(auto t: ima)
        {
            if(t == l)
            {
                flg = true;
                break;
            }
        }

        ima.push_back(l);
        cnt++;


    }

    cout << cnt << endl;

    return 0;
}