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>
#include <stdbool.h>
#define MAX 1000000

bool dbl(int *a, int len) {
  int i;
  for(i = 0; i < len-1; ++i) {
	if(a[i] == a[len-1]) { return true; }
  }
  return false;
}

int main(void)
{
  int s, len = 0, a[MAX];
  scanf("%d", &s);
  a[len] = s; ++len;
  while(true) {
    if(a[len-1]%2 == 0) { a[len] = a[len-1]/2; ++len; }
    else { a[len] = 3*a[len-1] + 1; ++len; }
    if(dbl(a, len)) { break; }
    if(len >= 1000000) { break; }
  }
  //int i;
  //for(i = 0; i < len; ++i) {
  //  printf(" %d", a[i]);
  //}
  //printf("\n");
  printf("%d\n", len);
  return 0;
}