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 <utility>
#include <functional>
#include <math.h>
#include <cmath>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <tuple>
#include <stdlib.h>
using namespace std;
#define rep(i,n,x) for(int i=n;i<x;i++)
#define rem(i,n,x) for(int i=n;i>x;i--)
#define SORT(x) sort(x.begin(),x.end())
#define REV(x) reverse(x.begin(),x.end())
#define all(x) x.begin(),x.end()
#define pb push_back
#define mp make_pair
#define setp(x) setprecision(x)
using LL = long long;
using ULL = unsigned long long;

int f(int n) {
	if (n % 2 == 0) return n / 2;
	else return 3 * n + 1;
}

int main() {
	int s;
	cin >> s;
	set<int> a;
	bool b = true;
	int n = s, ans = 0;
	while (1) {
		ans++;
		if (a.count(n)) {
			break;
		}
		else a.insert(n);
		n = f(n);
	}
	cout << ans << endl;
	return 0;
}