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)
ml=[0]*100001
l=[nil]
i=1
s,t=0,0
while i<=n
  if ml[x]>0
    s=ml[x]
    t=i-1
    break
  end
  ml[x]=i
  l << x
  i+=1
  x=x**2%m
end
if s==0
  puts l[1..n].sum
else
  c=l[1...s].sum
  c+=l[s..t].sum*((n-s+1)/(t-s+1))
  c+=l[s,((n-s+1) % (t-s+1))].sum
  puts c
end