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 Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)

import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long N = sc.nextLong();
		long X = sc.nextLong();
		long M = sc.nextLong();

		sc.close();

		long sum = 0;
		long amari = 0;
		Long[] before = new Long[100000000];
		for (int i = 0; i < N; i++) {
			List<Long> a = Arrays.asList(before);
			if (a.contains(X)) {
				int start = a.indexOf(X);
				List<Long> sub = a.subList(start, i);
				long sum2 = 0;
				for (Long l : sub) {
					sum2 += l;
				}
				sum = sum + ((N - i) / (i - start)) * sum2;
				amari = (N - i) % (i - start);
				X = sub.get(0);
				break;
			}
			sum = sum + X;
			before[i] = X;
			X = (X * X) % M;
		}
		for (int i = 0; i < amari; i++) {
			sum = sum + X;
			X = (X * X) % M;
		}
		System.out.print(sum);
	}

}