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.strip.split.map(&:to_i)
v = Array.new(m,0)
c = Array.new(m,0)
a = Array.new(m,0)
ans = 0
n.times do |i|
  break if x == 0
  ans += x
  y = x * x % m
  if v[x] == 0
    v[x] = y
    c[x] = i
    a[x] = ans
  else
    l = ans - a[x]
    s = i - c[x]
    z = (n - i - 1)/s
    ans += z * l
    (i+z*s+1).upto(n-1) do
      ans += y
      y = v[y]
    end
    p ans
    exit
  end
  x = y
end
p ans