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 <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); i++)
#define rep1(i,n) for (int i = 1; i <= (n); i++)
#define repf(i,a,b) for (int i = (a); i < (b); i++)
#define repi(i,a,b) for (int i = (a); i <= (b); i++)
#define repr(i,a,b) for (int i = (a); i >= (b); i--)
#define all(x) (x).begin(),(x).end()
using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vd = vector<double>;
using pii = pair<int, int>;
using vc = vector<char>;
using vvc = vector<vc>;
using vs = vector<string>;
using tiii = tuple<int, int, int>;

const int N = 1000000;

int main() {
  int s;
  cin >> s;
  
  vi vect;
  int x = s;
  while (count(all(vect),x)==0) {
    vect.push_back(x);
    if(x%2 == 0) {
      x /= 2;
    }
    else {
      x = x*3+1;
    }
  }
  
  cout << vect.size()+1 << endl;
}