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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int count = 0,stock[1000000],flag=0,answer=0,s=0;



void check(){
  for(int i=0;i<count;i++){
    for(int j=i+1;j<count;j++){
      if(i!=0 && j!=0){
        if(stock[i]==stock[j]){
          flag=1;answer = j;
        }
      }  
    }
  }
}

int saiki(int in){
  stock[count]=in;
  count++;
  check();
    if(flag>0){
    printf("%d",answer);return 0;
  }
  if(in%2 == 0){
    return saiki(in/2);
  }
  else{
    return saiki(3*in+1);
  }

}
int main(){
  scanf("%d",&s);
  for(int i=0;i<1000000;i++){
  stock[i]=0;
}
  saiki(s);
}