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)

#!/usr/bin python3
# -*- coding: utf-8 -*-

# ループを検索して計算量圧縮
# 0->1->...->st->st+1->...->st+lp-1-┐
#             └─────────────────────┘
#                       lp


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

seen = set([])
seqs = []
for i in range(m+1):
    seen.add(x)
    seqs.append(x)
    x = pow(x,2,m)
    if x in seen:
        break
start = seqs.index(x)
lp = len(seqs) - start

lpsum = sum(seqs[start:])
lptimes = (n-(start+lp)+(lp-1))//lp
ret_sum = lpsum * lptimes
n -= lp * lptimes
ret_sum += sum(seqs[:n])
ret_val = seqs[n-1]

print(ret_sum)
#print(ret_val)