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.chomp.split.map(&:to_i)

h = {}
v = []

si = ei = nil

a = x
n.times do |i|
  v.push(a)
  if h.key?(a)
    si = h[a]
    ei = i
    break
  end
  h[a] = i

  a = (a * a) % m
end

if si.nil?
  puts v.sum
else
  v0 = v[0...ei]
  s0 = v0.sum
  v1 = v[si...ei]
  x = n - ei
  y = v1.size

  c = x / y
  d = x % y

  s1 = v1.sum * c
  s2 = v1[0...d].sum

  puts s0 + s1 + s2
end