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: Runtime Error)

n,x,m = map(int,input().split())

cycles = [0 for i in range(m)]
cycle_num = 1
cycle = 1
ans_list = [0 for i in range(m)]
ans_list[0] = x
next_num = (x**2)%m

for i in range(m):
    cycle += 1
    if cycles[next_num] != 0:
        cycle_num = cycle - cycles[next_num]
        break

    ans_list[i+1] += ans_list[i] + next_num
    cycles[next_num] = cycle
    next_num = (next_num**2)%m

cycle_sum = ans_list[cycle]- ans_list[cycle-cycle_num]
ans = 0

if n < cycle:
    ans = ans_list[n-1]
else:
    ans  = ans_list[cycle]
    ans += ((n-cycle)//cycle_num) * cycle_sum
    rem = ((n-cycle)%cycle_num)
    ans += ans_list[cycle_num + rem] - ans_list[cycle_num]

print(ans)