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:
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^{8}.
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^{8}.
Output
For every case, print the length of the cycle, or the first number that strictly exceeds 10^{8}.
Observation
Take into account that the sequences usually reach fast a “short” cycle.
Input
1 5 8 0 5 0 10 11 3 7 3 6 1 999 100000000 433 805 215476 0 1 33333333
Output
3 1 1 35 150001002 490 3