Problem:
Score : 500 points

Problem Statement

Let us denote by f(x, m) the remainder of the Euclidean division of x by m.

Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M).
Find \displaystyle{\sum_{i=1}^N A_i}.

Constraints

1 \leq N \leq 10^{10}

0 \leq X < M \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N X M

Output

Print \displaystyle{\sum_{i=1}^N A_i}.

Sample Input 1

6 2 1001

Sample Output 1

1369

The sequence A begins 2,4,16,256,471,620,\ldots Therefore, the answer is 2+4+16+256+471+620=1369.

Sample Input 2

1000 2 16

Sample Output 2

6

The sequence A begins 2,4,0,0,\ldots Therefore, the answer is 6.

Sample Input 3

10000000000 10 99959

Sample Output 3

492443256176507

Solution in Fortran (GNU Fortran 9.2.1) (judge verdict: Accepted)

program sequence_sum
  implicit none
  integer(8) :: n, num, tmp, a(0:100000) = 0
  integer :: x, m, b(0:100000) = 0, period, offset, rem, i
  read(*,*) n, x, m
  tmp = m
  if (n <= 2 * m) then
    num = x
    rem = int(n)
    do i = 2, rem
      x = mod(int(x, 8) * x, tmp)
      if (x == 0) exit
      num = num + x
    end do
    write(*,'(i0)') num
    stop
  end if
  a(1) = x
  b(x) = 1
  do i = 2, m + 1
    x = int(mod(int(x, 8) * x, tmp))
    a(i) = a(i - 1) + x
    if (b(x) > 0) then
      offset = b(x) - 1
      period = i - offset - 1
      exit
    end if
    b(x) = i
  end do
  num = (n - offset) / period
  rem = int(n - offset - num * period)
  write(*,'(i0)') (a(offset + period) - a(offset)) * num + a(offset + rem)
end program sequence_sum