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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define all(a)a.begin(),a.end()
using ll=long long;
const int INF = 1<<30;
const ll INFll =1LL<<62;
const int mod= int(1e9)+7;
//const int mod=998244353;
using P = pair<int,int>;
using Pl= pair<ll,ll>;
using ld=long double;
using V=vector<int>;
using Vl=vector<ll>;
using VV=vector<vector<int>>;
using VVl=vector<vector<ll>>;

int main(){
    ll n,m,x,s=0;
    cin >>n>>x>>m;
    vector<Pl>memo(m);
    for (ll i=0; i <n; ++i) {
        if(memo[x].first){
            auto t=i-memo[x].first,c=(n-1-i)/t;
            i+=t*c;
            s+=(s-memo[x].second)*c;
        }
        else {
            memo[x]={i,s};
        }
        s+=x;
        x*=x;x%=m;
    }
    cout <<s<<endl;
}