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: Wrong Answer)

require "set"
n, x, m = gets.split.map(&:to_i)


r = x
res = x
r_set = Set.new
r_set << x
r_arr = [r]
ind = 0
(n-1).times do |i|
  ind = i
  r = r ** 2 % m
  res += r
  break if r_set.include?(r)
  r_set << r
  r_arr << r
end


if r == 1
  puts res + (n - ind - 1)
elsif r == 0
  puts res
else
  count = 0
  sum = 0
  r_arr.reverse.each do |rnum|
    count += 1
    sum += rnum
    break if r == rnum
  end
  res += ((n - r_arr.length - 1) / count) * sum

  de = (n - r_arr.length - 1) % count
  de.times do 
    r = r ** 2 % m
    res += r
  end
  puts res
end