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 scn = new Scanner(System.in);
        int n = scn.nextInt();
        int x = scn.nextInt();
        int m = scn.nextInt();
        int sum=x;
        int a=x;
        for(int i=1;i<n;i++)
        {
          if(a>m)
          {a = ((int)Math.pow(a,2))%m;}
          else 
          {a=(int)Math.pow(a,2);}
 
            sum +=a;
 
        }
        System.out.println(sum);
    }
}