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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long s = sc.nextInt();
		sc.close();
		int ans = 0;
		Set<Long> set = new HashSet<Long>();
		for (int i = 0; i < 1_000_000; i++) {
			long tmp = s % 2 == 0 ? (s / 2) : (3 * s + 1);
			if (set.contains(tmp)) {
				ans = i + 2;
				break;
			} else {
				set.add(tmp);
				s = tmp;
			}
		}
		System.out.println(ans);
	}

	public static boolean isPrime(long jud) {
		double root = Math.sqrt(jud);
		for (int i = 2; i <= root; i++) {
			if (jud % i == 0) {
				return false;
			}
		}
		return true;
	}

	// 配列要素数が1だと前触れなくバグるので注意！！！REりたくなければ要素数を保証するか例外処理を入れろ！！！
	public static long gcd(final long[] param) {
		final int len = param.length;
		long g = gcd(param[0], param[1]);
		for (int i = 1; i < len - 1; i++) {
			g = gcd(g, param[i + 1]);
		}
		return g;
	}

	public static long gcd(long a, long b) {
		// a > b にする(swap)
		if (a < b) {
			a = a ^ b;
			b = a ^ b;
			a = a ^ b;
		}
		// ユークリッドの互除法
		long r = -1;
		while (r != 0) {
			r = a % b;
			a = b;
			b = r;
		}

		return a; // b には r=0 の値が入るため、a を返す
	}

	public static long lcm(long m, long n) {
		return m * n / gcd(m, n);
	}

	public static long lcm(final long[] param) {
		final int len = param.length;
		long l = lcm(param[0], param[1]);
		for (int i = 1; i < len - 1; i++) {
			l = lcm(l, param[i + 1]);
		}
		return l;
	}

	public static int[] arrayInt(Scanner sc, int n) {
		int[] array = new int[n];
		for (int i = 0; i < n; i++) {
			array[i] = sc.nextInt();
		}
		return array;
	}

	public static long[] arrayLong(Scanner sc, int n) {
		long[] array = new long[n];
		for (int i = 0; i < n; i++) {
			array[i] = sc.nextLong();
		}
		return array;
	}

	public static double[] arrayDouble(Scanner sc, int n) {
		double[] array = new double[n];
		for (int i = 0; i < n; i++) {
			array[i] = sc.nextDouble();
		}
		return array;
	}
}