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)

n, x, m = gets.split(' ').map{|i| i.to_i}
 
z = 0
a = Array.new
a[0] = x
k_s = nil
k_e = 0
z = x
(n - 1).times do |i|
  k_e = i
  a_last = a[-1] ** 2 % m
  k_s = a.index(a_last)
  if k_s
    break
  end
  a.push a_last
  z += a_last
end
  
if k_s
  roops = ((n - k_e) / (k_e - k_s + 1)) 
  roop_sum = a[k_s..-1].sum
  lest_sum = a[k_s..(k_s + (n - k_s ) % (k_e - k_s + 1) -1)].sum
  z = z + roop_sum * roops + lest_sum
end
  
puts z