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)
dp=[]
dp[1]=x
sum=[]
sum[1] = x
(2..n).each do |i|
  p = dp[i-1]**2%m
  if dp.include?(p)
    sum[n]=sum[i-1]
    l = [*dp.index(p)+1..i].length
    sum[n]+=((n-i-1)/l)*dp[dp.index(p)..i].sum
    sum[n]+=dp[dp.index(p),((n-i+1)%l)].sum
    break
  end
  break sum[n] = sum[i-1] if p==0
  dp[i] = p
  sum[i] = sum[i-1] + dp[i]
end
p sum[n]