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 <bits/stdc++.h>
using namespace std;
#define pb push_back

int ncr(int n,int r){ int res=1; if(r>n-r)r=n-r; for(int i=0;i<r;i++) {  res*=n-i;  res/=i+1; } return res; }

int gcd(int a,int b){
    if(b==0) return a;
    else return gcd(b,a%b);
}
vector<int>v;
bool isPresent(int n){
    for(int i=0;i<v.size();i++){
        if(v[i]==n) return true;
    }
    return false;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t=1;
    //cin>>t;
    int sum=0;
    while(t--){
        int n;
        cin>>n;
        v.pb(n);
        if(n%2==0) n=n/2;
        else n=3*n+1;
        while(isPresent(n)==false){
            v.pb(n);
            if(n%2==0) n=n/2;
            else n=3*n+1;
        }
        cout<<v.size()+1<<endl;
    }
}