# Eventually periodic sequence

Tags:
Time Limit:  1 s      Memory Limit:   128 MB
Submission：1     AC：1     Score：99.94

## Description

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.

## Samples

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
1 3 6 6 369