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: Runtime Error)

n,x,m = gets.split.map(&:to_i)
looop = [x]
forth = []
l = nil
k = nil
app = Hash.new(false)
app[x] = true
m.times do
  a = (looop[-1]**2)%m
  if a == 0
    puts looop.sum
    exit
  elsif app[a] == true
    z = looop.index(a)
    if z != 0
        forth = looop.slice(0..(z-1))
        k = forth.size
    else
        forth = [0]
        k = 0
    end
    if z == 0 && looop.size == 1
        l = 1
    else
        looop.slice!(0..(z-1))
        l = looop.size
    end
    break
  else
    looop << a
    app[a] = true
  end
end
ans = forth.sum + ((n-k)/l)*looop.sum
if (n-k)%l==0
    puts ans
else
    puts ans+looop.slice(0..((n-k)%l-1)).sum
end