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)
hash = Hash.new(0)
cnt = 0
ans = 0
loopans = 0
loopcnt = 0
y = x
while (hash[y] == 0 || hash[y] == 1) && cnt <= n-1
  hash[y] += 1
  ans += y
  loopans += y if hash[y] == 2
  loopcnt += 1 if hash[y] == 2
  cnt += 1
  y = (y*y) % m
end
if cnt >= n
  puts ans
else
  ans += loopans * ((n-cnt) / loopcnt)
  rest = (n-cnt) % loopcnt
  rest.times do
    ans += y
    y = (y*y) % m
  end
  puts ans
end