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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <vector>
#include <iostream>
#include <map>
using namespace std;
using ll = int64_t;

int main() {
  ll n,x,m;
  cin >> n >> x >> m;
  map<ll,ll> AM;
  vector<ll> A;
  ll sum = 0;
  bool found_cycle = false;
  while(n>0) {
    A.push_back(x);
    sum += x;
    n--;
    if(!found_cycle) {
      if(AM.count(x) == 1) {
        ll cycle_length = A.size()-AM[x];
        ll cycle_sum = 0;
        for(ll i=A.size()-cycle_length; i<A.size(); i++) {
          cycle_sum += A[i];
        }
        //cerr << "CYCLE: A[" << A.size() << "] = A[" << AM[x] << "] = " << x << " length=" << cycle_length << " sum=" << cycle_sum << endl;

        ll amt = n/cycle_length;
        sum += amt*cycle_sum;
        n -= amt*cycle_length;
        found_cycle = true;
      } else {
        AM[x] = A.size();
      }
    }
    x = (x*x)%m;
  }
  cout << sum << endl;
}