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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
#define faster ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
typedef long long ll;
 
typedef vector<int> vi;;
typedef vector<ll> vll;

#define l() length()
#define sz() size()
 
#define b() begin()0
#define e() end()
#define all(x) x.begin(), x.end()
 
#define pb(i) push_back(i)
#define PB() pop_back()
 
 
typedef pair<int, int> pii;
typedef pair<ll,ll> pll;
#define mp make_pair
#define Fi first
#define Se second
 
const int maxx=100009;
 
// Happy Coding!



void solve(){
    int s;
    cin>>s;
    set<int> st;
    st.insert(s);
    
    int curr=s;
    int i=1;
    while(1){
        i++;
        curr=(curr&1)?3*curr+1:curr/2;
        if(st.find(curr)!=st.end()){
            cout<<i;
            return ;
        }    
        st.insert(curr);
    }
}

 
int main() {
    faster;
    int t=1;
    // cin>>t;
    
    while(t-->0){
        solve();
    }
	return 0;
}