## Problem C: Eventually periodic sequence

Given is a function *f: 0..N --> 0..N* for a non-negative
*N* and a non-negative integer *n* ≤ *N*. One
can construct an infinite sequence *F = f*^{ 1}(n),
f^{ 2}(n), ... f^{ k}(n) ... , where
*f*^{ k}(n) is defined recursively as follows:
*f*^{ 1}(n) = f(n) and *f*^{ k+1}(n) =
*f(f*^{ k}(n)).
It is easy to see that each such sequence *F* is eventually
periodic, that is periodic from some point onwards, e.g 1, 2, 7, 5, 4,
6, 5, 4, 6, 5, 4, 6 ... .
Given non-negative integer *N ≤ 11000000 *, *n ≤ N* and
*f*, you are to compute the period of sequence *F*.

Each line of input contains *N*, *n* and the a
description of *f* in postfix notation, also known as Reverse
Polish Notation (RPN). The operands are either unsigned integer
constants or *N* or the variable *x*. Only binary
operands are allowed: + (addition), * (multiplication) and % (modulo, i.e. remainder of integer division).
Operands and operators are separated by whitespace. The operand %
occurs exactly once in a function and it is the last (rightmost, or
topmost if you wish) operator and its second operand is always *N*
whose value is read from input. The following function:

2 x * 7 + N %

is the RPN rendition of the more familiar infix `(2*x+7)%N`. All input
lines are shorter than 100 characters. The last line of input has
*N* equal 0 and should not be processed.
For each line of input, output one line with one integer number, the period
of *F* corresponding to the data given in the input line.

### Sample input

10 1 x N %
11 1 x x 1 + * N %
1728 1 x x 1 + * x 2 + * N %
1728 1 x x 1 + x 2 + * * N %
100003 1 x x 123 + * x 12345 + * N %
0 0 0 N %

### Output for sample input

1
3
6
6
369

*Piotr Rudnicki (from folklore, sometimes attributted to R. Floyd)*