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 PyPy3 (7.3.0) (judge verdict: Runtime Error)

import sys
stdin = sys.stdin
sys.setrecursionlimit(10**6)
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
nn = lambda: list(stdin.readline().split())
ns = lambda: stdin.readline().rstrip()

n,x,m = na()
from collections import defaultdict
d = defaultdict(int)
a = x
d[a] = 1
dp = [0,a]

for i in range(2,n+1):
    b = (a*a)%m
    if b in d:
        j = d[b]
        loo = n-j-1
        ll = i-j
        d = loo//ll
        e = loo%ll
        loosum = dp[i-1]-dp[j-1]
        print(loosum*d+dp[j+e+1 if j+e+1 != ll else loosum])
        exit()
    d[b] = i
    dp.append(dp[-1]+b)
    a = b

print(dp[-1])