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)

#include<bits/stdc++.h>
using namespace std;

int intpow(int x,int y){
  double a=x,b=y;
  return pow(a,b);
}//(x,y) xをy乗する

int intfabs(int x){
  double a=x;
  return fabs(a);
}//(x) の絶対値

int in(){
  int a;
  cin >> a;
  return a;
}//int型関数の受け取り

void forin(int* x,int y){
  for (int i=0;i<y;i++){
    cin >> x[i];
  }
}//配列の受け取り

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

main(){
  int c[500000],a,s=in();
  a=s;
  for (int i=0;i<500000;i++) c[i]=0;
  for (int i=1;true;i++){
    a=fn(a);
    c[a]++;
    if (c[a]==2){
      cout << i+1 << endl;
      return 0;
    }
  }
}