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: Runtime Error)

#include <iostream>
#include <vector>
using namespace std;
using ll = int64_t;
#define rep(i, j, n) for (int i = j; i < (int)(n); ++i)

int main() {
  ll n, x, m;
  cin >> n >> x >> m;
  vector<bool> used(m * 2);
  vector<ll> one_cycle;

  int start = 0;
  ll now = x;
  rep(i, 0, n) {
    used[now] = true;
    one_cycle.push_back(now);
    now = now * now % m;
    if (used[now]) {
      rep(j, 0, one_cycle.size()) if (one_cycle[j] == now) {
        start = j;
        break;
      }
      break;
    }
  }

  ll s1 = 0, s2 = 0, loop = one_cycle.size() - start;
  rep(i, 0, start) s1 += one_cycle[i];
  rep(i, start, one_cycle.size()) s2 += one_cycle[i];
  n -= start;
  ll ans = s1 + n / loop * s2;
  rep(i, 0, n % loop) {
    ans += now;
    now = now * now % m;
  }
  cout << ans << endl;
  return 0;
}