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())
mn = min(n, m)
P = [] # (x,pre_sum)
sum_p = 0 # sum of pre + cycle
X = [-1] * m # for cycle check
for i in range(mn):
    if X[x] > -1:
        cyc_len = len(P) - X[x]
        pre = P[X[x]]
        cyc = (sum_p - pre) * ((n - X[x]) // cyc_len)
        nxt = P[X[x] + (n - X[x]) % cyc_len] - P[X[x]]
        print(pre + cyc + nxt)
        exit()
    P.append(sum_p)
    sum_p += x
    X[x] = i
    x = x*x % m
print(sum_p)