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 <cstdio>
#include <cstdint>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <array>
#include <vector>
#include <unordered_map>
#include <unordered_set>

int main()
{
    int s;
    std::cin >> s;

    std::unordered_set<int> vals;

    vals.insert(s);
    int prev = s;
    for (int i = 2; i < 1000000; ++i) {
        int f = 0;
        if (prev % 2 == 0) {
            f = prev / 2;
        } else {
            f = 3 * prev + 1;
        }

        if (vals.count(f) != 0) {
            std::cout << i << std::endl;
            break;
        }
        vals.insert(f);
        prev = f;
    }

    return 0;
}