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 Ruby (2.7.1) (judge verdict: Accepted)

N, X, M = $stdin.gets.split.map {|i| i.to_i }
case X
when 0
  puts "0"
  exit
when 1
  puts N
  exit
end
A = { 0 => X }
R = { X => 0 }
last = 0
sum0 = X
offset = nil
1.upto(N - 1) do |i|
  f = (A[i - 1] ** 2) % M
  if f == 0
    puts sum0
    exit
  elsif offset = R[f]
    break
  end
  A[i] = f
  R[f] = i
  last = i
  sum0 += f
end
offset ||= 0
length = last + 1 - offset
sum = 0
0.upto(offset - 1) do |i|
  sum += A[i]
end
0.upto(last - offset) do |i|
  a = A[offset + i]
  c = (N - offset - i + length - 1) / length
  sum += a * c
end
puts sum