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: Wrong Answer)

import sys
N , X , M = map(int,input().split())
flag = [0]*M
S = X
SN = [0]*(M+1)
SN[1] = X 
A = X
if M == 1:
  print(0)
  sys.exit()
if X == 0:
  print(0)
  sys.exit()
if A%M == (A**2)%M:
  print(A*N)
  sys.exit()
loop = 1
flag[X%M] = 1
for i in range(N - 1):
  A = (A**2)%M
  if flag[A] == 0:
    flag[A] = loop
  else:
    loops = flag[A]
    break
  S += A
  SN[loop+1] = S
  loop += 1
if loop <N:
  ans = 0
  ans += SN[loops]
  dS = SN[loop] - SN[loops]
  dl = loop - loops
  ans += ((N-loops)//dl)*dS
  ans += SN[(N-loops)%dl+loops] - SN[loops]
  print(ans)
else:
  print(S)