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: Wrong Answer)

//テンプレ
#include <bits/stdc++.h>
//#define int long long
#define REP(i,n) for(int i = 0; i < (n); ++i)
#define FOR(i,a,b)  for(int i = (a); i <= (b); ++i)
#define ALL(v) (v).begin(),(v).end()

using namespace std;
//constexpr int MOD = 10e9+7;
//constexpr int INF = 1LL << 60;
template <class T> inline bool chmin(T& a, T b){if(a > b){ a = b; return 1;} return 0;}  
template <class T> inline bool chmax(T& a, T b){if(a < b){ a = b; return 1;} return 0;}  
const int dx[] = {0, 1, 0, -1, 1, -1, 1, -1}, dy[] = {1, 0, -1, 0, 1, -1, -1, 1};
//

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

signed main (){
   ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
   int s;
   cin >> s;
   int a = s;
   int n = 1;
   int ans;
    while(true)
    {
        if( a == 1 || a == 2 || a== 4)
        {
            cout << n +2 <<endl;
            return 0;
        }
        a=f(a);
        n++;
    }
}