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: Runtime Error)

import java.util.*;
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();
        //long N = sc.nextLong();

        long res=X;
        long A=X;

        boolean[] ch=new boolean[(int)M];
        long[] rui = new long[(int)M+1];
        int[] ban = new int[(int)M];
        
        rui[1] = X;
        ban[(int)X] = 1;
        ch[(int)X]=true;

        for(int i=2; i<=Math.min(M,N); i++){
            A = (long)Math.pow(A, 2)%M;
            ban[(int)A]=i;
            rui[i] = rui[i-1] + A;
            if(ch[(int)A]){
                long junkann = rui[i-1] - rui[ban[(int)A-1]];
                long junkannkazu = (N-i+1)/(i-ban[(int)A]);
                long junkannMod = (N-i+1)%junkannkazu;
                res = rui[i-1] + junkann*junkannkazu + rui[(int)(i-1-i+ban[(int)A]+junkannMod)] - rui[ban[(int)A-1]];
                System.out.println(res);
                return;
            }
            
        }
        
        System.out.println(rui[(int)N]);
        }
    }