//callatz假设的1000以内的数验证

#include <iostream>

using namespace std;

int main()
{
int n;
cin >> n;
int i = 0;
while(n!=1)
{

i++;
if(n%2==1)
{
n = (3 * n + 1)/2;
}
else
{
n = n / 2;
}
//cout << n << endl;
}
cout << i << endl;
return 0;
}

 

//callatz假设的1000以内的数验证