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)

from collections import defaultdict
dic = defaultdict(int)
N, X, M = map(int, input().split())
A = [X]
ans = X
fir = -1
if N>1:
    temp = 1
    dic[X]=temp
    for i in range(2, N+1):
        a = A[-1]*A[-1]%M
        if dic[a]>0:
            fir = dic[a]
            break
        A.append(a)
        ans+=a
        temp+=1
        dic[a]=temp
    if fir!=-1:
        T = temp-fir+1#周期
        Tsum = 0 #周期の和
        TempSum = 0 #後半あまりの和
        M=N-fir+1 # 先頭のぞく 
        amari = M%T # 後半あまり
        for k in dic:
            if dic[k]>=fir and dic[k]<=temp:
                Tsum+=k
            if dic[k]>=fir and dic[k]<fir+amari:
                TempSum += k
        ans+=TempSum
        ans+=Tsum*(M//T-1)
    print(ans)
else:
    print(ans)