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 <stdlib.h>

#define MMAX 1000000

typedef unsigned int u32;
typedef unsigned char u8;

static u32 map[MMAX];

u32 f(u32 n) {
  if (n % 2 == 0) {
    return n / 2;
  } else {
    return 3*n + 1;
  }
}

u32 max(u32 a, u32 b) {
  return (a < b) ? b : a;
}

int main() {
  u32 s;
  scanf("%u", &s);
  for (u32 i = 1; ;) {
    map[s] = i++;
    s = f(s);
    if (map[s] != 0) {
      printf("%u\n", max(map[s], i));
      return 0;
    }
  }
  return 0;
}