# 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 = f1(n), f2(n), . . . fk(n) . . ., where fk(n) is defined recursively as follows: f 1 (n) = f(n) and fk+1(n) = f(fk(n)). It is easy to see that each such se- quence 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. Input 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: 2x7+N% is the RPN rendition of the more familiar infix ‘(2*x+7)%N’. All input lines are shorter than 100 char- acters. The last line of input has N equal ‘0’ and should not be processed. Output 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 % 000N% Sample Output 1 3 6

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