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 = gets.split.map(&:to_i)
memo = Array.new(10 ** 5 + 1, false)
tmp_nums = []
until memo[x]
  memo[x] = true
  tmp_nums << x
  x **= 2
  x %= m
end

tmp_nums_size = tmp_nums.size
first_x_index = tmp_nums.find_index(x)
not_loop_nums = tmp_nums[0, first_x_index]
not_loop_nums_size = not_loop_nums.size
loop_nums = tmp_nums[first_x_index..-1]
loop_nums_size = loop_nums.size

if n <= tmp_nums_size
  puts tmp_nums[0, n].sum
else
  ans = tmp_nums.sum
  rest = n - tmp_nums_size
  loop_count, mod = rest.divmod(loop_nums_size)
  ans += loop_nums.sum * loop_count
  ans += loop_nums[0, mod].sum
  puts ans
end