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)

opens = Hash.new()
accum = [0]
cnt = 0
a = x

while !opens.include?(x)
  accum << a
  opens[x] = cnt
  x = (x * x) % m
  a += x
  cnt += 1
end

pathlen = opens[x]
pathaccum = accum[pathlen]
loopsize = cnt - pathlen
loopaccum = accum[cnt] - accum[pathlen]
loopstart = x


res = 0
if n <= pathlen
  res = accum[n]
else
  t = (n - pathlen) / loopsize
  rest = n - pathlen - t * loopsize

  a = pathaccum + loopaccum * t
  x = loopstart
  rest.times do |i|
    a += x
    x = x * x % m
  end

  res = a
end

p res