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 Python (3.8.2) (judge verdict: Accepted)

n,x,m=map(int, input().split())
alist=[]
blist=[0]*m
num=x%m
while blist[num]==0:
  alist.append(num)
  blist[num]=1
  num=(num**2)%m
num1=alist.index(num)
num2=len(alist)
if x==1:
  print(n)
elif n<=num1:
  print(sum(alist[0:n]))
else:
  cycle=(n-num1)//(num2-num1)
  print(sum(alist[0:num1])+cycle*sum(alist[num1:num2+1])+sum(alist[num1:n-cycle*(num2-num1)]))