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)

#define _USE_MATH_DEFINES
#include <cassert>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <algorithm>
#include <complex>
#include <cmath>
#include <numeric>
#include <bitset>
#include <functional>

using namespace std;

#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}

typedef long long int64;
typedef pair<int, int> ii;
const int INF = 1 << 29;
const int MOD = 1e9 + 7;

const int N = 1e6 + 10;
int dp[N];

int main() {
  int s;
  scanf("%d", &s);
  memset(dp, 255, sizeof(dp));
  int k = 0;
  dp[s] = 0;
  int ret = -1;
  for (++k; ; ++k) {
    if (s % 2 == 0) {
      s /= 2;
    } else {
      s = s * 3 + 1;
    }
    if (dp[s] >= 0) {
      ret = k;
      break;
    }
    dp[s] = k;
  }
  printf("%d\n", ret + 1);
  return 0;
}