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)
a = x
seen = {}
sum = 0

i = 0
start = nil
nums = []
while i < n
  nums << a
  if seen[a]
    start = seen[a]
    break
  end
  sum += a
  seen[a] = i
  i += 1
  a = (a * a) % m
end
if i == n
  puts sum
  exit
end
len = i - start

rangesum = 0
(start...i).each do |j|
  rangesum += nums[j]
end

slots = (n - i) / len
r = (n - i) % len
total = sum + rangesum * slots
r.times do
  total += a
  a = (a * a) % m
end

puts total