Collatz pseudo-sequences (1)

Let us define sequences similar to those of Collatz with two parameters $x$ and $y$ . Given a number $n$ , the algorithm to get the next number is:

if $n$ is even, we move to $n/2 + x$ ;
otherwise, we move to $3n + y$ .

The standard Collatz sequence corresponds to $x = 0$ and $y = 1$ .

Given $x$ , $y$ and a starting number $n$ , compute the length of the cycle reached by applying the above algorithm. For example, if $x = 1$ , $y = 5$ and $n = 8$ , then the defined sequence is 8, 5, 20, 11, 38, 20, 11, 38, … so the cycle has length 3.

Since numbers can become very large, and we have no mathematical guarantee that we will reach a cycle, we will stop if at some point the sequence reaches a number greater than $10^6$ .

Input

Input consists of several cases, each with three natural numbers $x$ , $y$ and $n$ . Assume that both $x$ and $y$ do not exceed 1000, that $y$ is odd (for the sequence to have some interest), and that the initial $n$ is not larger than $10^6$ .

Output

For every case, print the length of the cycle, or the first number that strictly exceeds $10^6$ .

Problem information

Author: Unknown
Translator: Salvador Roura

Generation: 2026-01-25T11:54:11.940Z