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>
//#include<atcoder/all>

using namespace std;
//using namespace atcoder;


int main(){
    long long N,X,M;
    cin>>N>>X>>M;
    set<long long> s;
    long long A=X;
    s.insert(A);
    long long sum=A;
    while(N-1>0){
        A=A*A%M;
        if(s.find(A)!=s.end()){
            long long sum2=A;
            long long c=1;
            long long A2=A;
            while(1){
                A2=A2*A2%M;
                if(A2==A) break;
                c++;
                sum2+=A2;
                //cout<<c<<","<<sum2<<endl;
            }
            sum+=sum2*((N-1)/c);
            cout<<N<<","<<c<<endl;
            N=(N-1)%c;
            s.clear();
            if(N==0) break;
            N++;
        }
        s.insert(A);
        sum+=A;
        //cout<<A<<endl;
        N--;
    }
    cout<<sum<<endl;
}