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 <iostream>
#include <map>
#include <vector>

using namespace std;
typedef long long ll;
int main() {
  ll n, x, m;
  cin >> n >> x >> m;
  map<ll, int> mp;
  int idx = 0;
  vector<ll> path;
  do {
    mp[x] = idx++;
    path.emplace_back(x);
    x = x * x % m;
  } while (!mp.count(x));
  int start = mp[x];
  int len = idx - start;
  ll sum = 0;
  for (int i = start; i < idx; ++i)
    sum += path[i];
  ll ans = 0;
  if (n < start) {
    for (int i = 0; i < n; ++i)
      ans += path[i];
  } else {
    for (int i = 0; i < start; ++i)
      ans += path[i];
    ll loop = (n - start) / len;
    ans += loop * sum;
    ll rem = (n - start) % len;
    for (int i = 0; i < rem; ++i)
      ans += path[start + i];
  }
  cout << ans;
}