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;
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<vi> vvi;
typedef vector<vii> vvii;
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define eb emplace_back
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define SZ(x) (int)(x).size()
#define endl '\n'
#define FOR(a, b, c) for (auto a = (b); (a) < (c); ++(a))
#define F0R(a, b) FOR (a, 0, (b))
#define CEIL(a, b) ((a) + (b)-1) / (b)
template <class T>
bool ckmin(T& a, const T& b) {
  return a > b ? a = b, true : false;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, true : false;
}
#ifndef DEBUG
#define DEBUG 0
#endif
#define dout \
  if (DEBUG) std::cerr
#define dvar(...) " [" << #__VA_ARGS__ ": " << mt(__VA_ARGS__) << "] "
template <typename T>
struct IsC {
  template <typename C>
  static char test(typename C::const_iterator*);
  template <typename C>
  static int test(...);
  static const bool value = sizeof(test<T>(0)) == sizeof(char);
};
template <>
struct IsC<string> {
  static const bool value = false;
};
template <typename C>
typename enable_if<IsC<C>::value, ostream&>::type operator<<(ostream&, const C&);
template <typename T1, typename T2>
ostream& operator<<(ostream&, const pair<T1, T2>&);
template <typename... T> using TS = tuple_size<tuple<T...>>;
template<size_t idx, typename... T>
typename enable_if<TS<T...>::value == idx + 1, ostream&>::type
operator<<(ostream& o, const tuple<T...>& t) {
  return o << ", " << get<idx>(t) << ')';
}
template<size_t idx, typename... T>
typename enable_if<0 < idx && idx + 1 < TS<T...>::value, ostream&>::type
operator<<(ostream& o, const tuple<T...>& t) {
  return operator<<<idx + 1>(o << ", " << get<idx>(t), t);
}
template<size_t idx = 0, typename... T>
typename enable_if<1 < TS<T...>::value && !idx, ostream&>::type
operator<<(ostream& o, const tuple<T...>& t) {
  return operator<<<idx + 1>(o << '(' << get<idx>(t), t);
}
template<typename T>
ostream& operator<<(ostream& o, const tuple<T>& t) {
  return o << get<0>(t);
}
template <typename T1, typename T2>
ostream& operator<<(ostream& o, const pair<T1, T2>& p) {
  return o << mt(p.fi, p.se);
}
template <typename C>
typename enable_if<IsC<C>::value, ostream&>::type operator<<(
    ostream& o, const C& c) {
  o << '[';
  for (auto it = c.cbegin(); it != c.cend(); ++it)
    o << *it << (next(it) != c.cend() ? ", " : "");
  return o << ']';
}
template <typename T>
void tprint(vector<vector<T>>& v, size_t width = 0, ostream& o = cerr) {
  if (!DEBUG) return;
  for (const auto& vv : v) {
    for (const auto& i : vv) o << setw(width) << i;
    o << endl;
  }
}

int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0); cin.tie(0);

  ll n, x, m; cin >> n >> x >> m;
  vector<ll> dp;
  vi first(m, -1);
  first[x] = 0;
  dp.pb(x);
  ll sum = x;
  for (int i = 1;; ++i) {
    x = x * x % m;
    sum += x;
    if (~first[x]) {
      ll offset = first[x];
      ll loopsz = i - first[x];
      if (n <= i) {
        cout << dp[n - 1] << endl;
      } else {
        ll sol = 0;
        if (offset) {
          sol += dp[offset - 1];
        }
        ll offsetval = sol;
        ll loopval = sum - dp[first[x]];
        n -= offset;
        sol += loopval * (n / loopsz);
        if (n % loopsz) {
          sol += dp[offset - 1 + n % loopsz] - offsetval;
        }
        cout << sol << endl;
      }
      return 0;
    }
    first[x] = i;
    dp.pb(sum);
  }
  
  return 0;
}