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 <map>
#include <set>
#include <vector>
#define REP(i, n) for (int i = 0; i < (n); i++)
using namespace std;

void solve(long long s){
    int count = 0;
    if (s == 1) count = 0;
    else if (s == 2) count = 0;
    else {
        while (s != 4) {
            if (s % 2) s = 3 * s + 1;
            else s /= 2;
            count++;
        }
    }
    cout << count + 4<< '\n';
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    long long s;
    scanf("%lld",&s);
    solve(s);
    return 0;
}