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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
int main(){
	long long n,x,m;
	cin>>n>>x>>m;
	set<long long> s;
	vector<long long> v;
	while(1){
		if(s.find(x)!=s.end())break;
		s.insert(x);
		v.push_back(x);
		x=x*x;
		x%=m;
	}
	long long cnt=0;
	if(v.size()>=n){
		for(int i=0;i<n;i++){
			cnt+=v[i];
		}
		cout<<cnt<<endl;
		return 0;
	}
	int j;
	for(int i=0;i<v.size();i++){
		if(v[i]==x){
			j=i;
			break;
		}
		cnt+=v[i];
	}
	long long ans=0;
	if(j==0){
		for(int i=0;i<v.size();i++){
			ans+=v[i];
		}
		ans=ans*(n/(long long)(v.size()));
		int tmp=n%(long long)(v.size()); 
		if(tmp!=0){
			for(int i=0;i<=tmp;i++){
				ans+=v[i];
			}
		}
		cout<<ans<<endl;
		return 0;
	}
	for(int i=j;i<v.size();i++){
		ans+=v[i];
	}
//	cout<<1;
	ans=ans*((n-(long long)(j))/(long long)(v.size()-j));
	long long k=(n-(long long)(j))%(long long)(v.size()-j);
//	cout<<1;
	if(k>0){
		for(int i=j;i<j+k;i++){
			ans+=v[i];
		}
	}
	cout<<cnt+ans<<endl;
	return 0;
}