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: Accepted)

#include<stdio.h>

int main(void){

  int a[1000000] = {0};
  scanf("%d", &a[0]);       //入力のsは初項にあたる

  int ans = 0;                  //答え

  for(int i = 0 ; i < 1000000 ; i++){

    if( a[i] % 2 == 0 ){      //nが偶数の時
      a[i + 1] = a[i] / 2;     
    }else{                       //nが奇数の時
      a[i + 1] = 3 * a[i] + 1;
    }

    for(int j = 0 ; j < i ; j++){

      if(a[i + 1] == a[j]){
	ans = i + 2;
	break;
      }
      
    }

    if(ans != 0)     break;   //もう答えが出たなら先をやらない

    
  }


  printf("%d", ans);
	
  
  return 0;
}