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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

// typedef
// -----------------------------------------------------------------------------
typedef long long ll;
typedef long double ld;

// container util
// -----------------------------------------------------------------------------
#define ALL(c)  (c).begin(),(c).end()
#define SZ(c) ((ll)(c).size())
#define MAX(c) *max_element(ALL(c))
#define MIN(c) *min_element(ALL(c))
#define SORT(c) sort((c).begin(),(c).end())

// repetition
// -----------------------------------------------------------------------------
#define FOR(i, a, b) for(ll i= (a); i < (ll)(b); ++i)
#define REP(i, n) FOR(i, 0, n)

// constant
// -----------------------------------------------------------------------------
const ll INF = 1e9;
const ll MOD = 1e9 + 7;
const double EPS = 1e-9;
const double PI = acos(-1.0);

// debug
// -----------------------------------------------------------------------------
#define DUMP(x)  cerr << #x << " = " << (x) << endl


int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(15);
    int s; cin >> s;
    bool A[1000000];
    int ans = 1;
    while(!A[s]) {
        ans++;
        A[s] = true;
        if(s % 2 == 0) {
            s /= 2;
        } else {
            s = 3 * s + 1;
        }
    }
    cout << ans << "\n";
}