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())

s = set()
idx = {i: -1 for i in range(m)}
cumsum = [0]
val = x
idx[val] = 0
for i in range(n):
    cumsum.append(cumsum[-1] + val)
    val = pow(val, 2, m)
    if idx[val] != -1:
        break
    idx[val] = i + 1

if len(cumsum) > n:
    print(cumsum[n])
else:
    loop_len = i + 1 - idx[val]
    print((cumsum[i+1]-cumsum[idx[val]]) * ((n-idx[val])//loop_len) + cumsum[idx[val]+(n-idx[val])%loop_len])