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]
h = {}
h[x] = 0
t = 0
n.pred.times do |i|
  t = a[i] * a[i] % m
  break if h.key?(t)
  h[t] = 0
  a << t
end
b = a.dup
a = a[0...a.find_index{ _1 == t }]
b -= a
ans = a.sum
if b.size > 0
  ans += b.sum * ((n - a.size) / b.size)
  ans += b[0, (n - a.size) % b.size].sum
end
puts ans