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: Wrong Answer)

import java.util.Scanner;

public class Main {

	static long mod = 1000000007;

	static double function(double x, long m) {
		return x%m;
	}

	public static void main(String[] args) {
		// AtCoder_E
		Scanner sc = new Scanner(System.in);
		long N = sc.nextLong();
		int X = sc.nextInt();
		int M = sc.nextInt();
	    // Scannerクラスのインスタンスをクローズ
		sc.close();


		double ans = X;
		double before = X;

		for(int i=1; i<N; i++) {
			before = function(Math.pow(before,2),M);
			ans += before;
		}

		System.out.println(ans);

	}
}