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 C# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Collections;
using System.Collections.Specialized;
using System.Linq;
using System.Text;
using System.IO;
using System.Reflection;
using static System.Math;
using System.Numerics;
using System.Threading;
static class Program{
	const int mod=(int)1e9+7;
	const double eps=1e-11;
	static void Main(){
		Sc sc=new Sc();
		var s=sc.La;
		var h=new long[(int)s[2]+1];
		var d=new long[(int)s[2]+2];
		long p=s[1],ans=p;
		d[p]=ans;
		h[p]=1;
		var g=s[0];
		for(int i = 2;i<=s[0];i++) {
			p*=p;
			p%=s[2];
			if(p==0){break;}
			else if(h[p]!=0){
				g-=i;
				var m=i-h[p];
				ans+=(ans-d[p]+p)*(g/m);
				g-=g/m*m;
				for(int k = (int)g;k>0;k--) {
					p*=p;
					p%=s[2];
					ans+=p;
				}
				break;
			}
			h[p]=i;
			ans+=p;
			d[p]=ans;
		}
		Console.WriteLine("{0}",ans);
	}
}
public class Sc{
	public int I{get{return int.Parse(Console.ReadLine());}}
	public long L{get{return long.Parse(Console.ReadLine());}}
	public double D{get{return double.Parse(Console.ReadLine());}}
	public string S{get{return Console.ReadLine();}}
	public int[] Ia{get{return Array.ConvertAll(Console.ReadLine().Split(),int.Parse);}}
	public long[] La{get{return Array.ConvertAll(Console.ReadLine().Split(),long.Parse);}}
	public double[] Da{get{return Array.ConvertAll(Console.ReadLine().Split(),double.Parse);}}
	public string[] Sa{get{return Console.ReadLine().Split();}}
	public object[] Oa{get{return Console.ReadLine().Split();}}
	public int[] Ia2{get{return Array.ConvertAll(("0 "+Console.ReadLine()+" 0").Split(),int.Parse);}}
	public int[] Ia3(string a,string b){return Array.ConvertAll((a+Console.ReadLine()+b).Split(),int.Parse);}
	public int[] Ia3(int a){return Array.ConvertAll((Console.ReadLine()+" "+a.ToString()).Split(),int.Parse);}
	public long[] La2{get{return Array.ConvertAll(("0 "+Console.ReadLine()+" 0").Split(),long.Parse);}}
	public long[] La3(string a,string b){return Array.ConvertAll((a+Console.ReadLine()+b).Split(),long.Parse);}
	public long[] La3(int a){return Array.ConvertAll((Console.ReadLine()+" "+a.ToString()).Split(),long.Parse);}
	public double[] Da2{get{return Array.ConvertAll(("0 "+Console.ReadLine()+" 0").Split(),double.Parse);}}
	public double[] Da3(string a,string b){return Array.ConvertAll((a+Console.ReadLine()+b).Split(),double.Parse);}
	public T[] Arr<T>(int n,Func<T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f();}return a;}
	public T[] Arr<T>(int n,Func<int,T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f(i);}return a;}
	public T[] Arr<T>(int n,Func<string[],T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f(Console.ReadLine().Split());}return a;}
	public T[] Arr<T>(int n,Func<int,string[],T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f(i,Console.ReadLine().Split());}return a;}
}