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.io.*;
import java.util.*;

class Main {
	public static void main(String[] $) {
		new Main().run();
	}
	
	final long M=998244353;
	
	void run() {
		Scanner sc=new Scanner(System.in);
		int N=sc.nextInt();
		int K=sc.nextInt();
		long[] a=new long[N+1];
		int[] L=new int[K];
		int[] R=new int[K];
		for (int i=0;i<K;++i) {
			L[i]=sc.nextInt();
			R[i]=sc.nextInt();
		}
		a[0]=1;
		for (int i=1;i<=N;++i){
			a[i]=a[i-1];
			for (int j=0;j<K;++j) {
				if (i-L[j]>=0)   a[i]=(a[i]+a[i-L[j]])%M;
				if (i-R[j]-1>=0) a[i]=(a[i]+M-a[i-R[j]-1])%M;
			}
		}
		System.out.println((a[N-1]-a[N-2]+M)%M);
	}
	
	void tr(Object...objects) {System.out.println(Arrays.deepToString(objects));}
}