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 FOR(i, m, n) for (int i = m; i < (n); i++)
#define RFOR(i, m, n) for (int i = (m - 1); i >= 0; i--)
#define REP(i, n) FOR(i, 0, n)
#define RREP(i, n) RFOR(i, n, 0)
#define ALL(v) v.begin(), v.end()
#define RALL(v) v.rbegin(), v.rend()
#define print(ele) cout << (ele) << endl
#define print10(ele) cout << fixed << setprecision(10) << (ele) << endl

using namespace std;
typedef long long ll;
typedef vector<int> ivec;
typedef vector<string> svec;
typedef vector<ll> lvec;

const int mod = 1e9 + 7;
const ll INF = 1000000000000000000LL;

int f(int n) {
  if (n % 2)
    return 3 * n + 1;
  else
    return n / 2;
}

int main() {
  int s;
  set<int> a;
  cin >> s;
  a.insert(s);
  int count = 1;
  while (1) {
    count++;
    int sb = a.size();
    s = f(s);
    a.insert(s);
    int sa = a.size();
    if (sb == sa) {
      print(count);
      return 0;
    }
  }
  return 0;
}