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: Accepted)

import bisect, copy, heapq, math, sys
from collections import *
from functools import lru_cache
from itertools import accumulate, combinations, permutations, product
def input():
    return sys.stdin.readline()[:-1]
def ruiseki(lst):
    return [0]+list(accumulate(lst))
def celi(a,b):
    return -(-a//b)
sys.setrecursionlimit(5000000)
mod=pow(10,9)+7
al=[chr(ord('a') + i) for i in range(26)]
direction=[[1,0],[0,1],[-1,0],[0,-1]]

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


cnt=1
now=x
ans=x
jisyo=defaultdict(list)
jisyo[now]=[ans,cnt]
while cnt<n:
    nex=now**2%m
    cnt+=1
    ans+=nex
    now=nex
    if nex in jisyo:
        tmp=jisyo[nex]
        # print(tmp)
        ans+=(ans-tmp[0])*((n-cnt)//(cnt-tmp[1]))
        cnt=n-(n-cnt)%(cnt-tmp[1])
        break
    else:
        jisyo[now]=[ans,cnt]
    # print(ans,cnt,now)
for i in range(n-cnt):
    now=now**2%m
    ans+=now
print(ans)