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.chomp.split(' ').map(&:to_i)
num_list = [x]
i = 0
num = x
while true
  a = (num**2) % m
  num_list << a
  num = num_list[-1]
  if num_list.uniq.size != num_list.size
    break
  end
end
rep_id = -1
num_list.each.with_index do |i,id|
  if i == num
    rep_id = id
    break
  end
end
ans = 0
if n <= num_list.size - 1
  puts num_list[0..(n-1)].sum
else
  ans += num_list[0..(rep_id-1)].sum
  n -= rep_id - 1
  size = num_list.size - rep_id - 1
  ans += num_list[rep_id..(num_list.size - 2)].sum * (n/size)
  ans += num_list[rep_id..(rep_id +(n%size)) - 2].sum
  puts ans
end