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 <algorithm>
#include <cmath>
#include <deque>
#include <iostream>
#include <iterator>
#include <limits>
#include <map>
#include <set>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;
using i64 = int64_t;
#define rep(i, s, n) for (i64 i = (s); i < (n); ++i)

void run() {
  i64 n, x, m;
  cin >> n >> x >> m;
  vector<i64> i_pre(m, -1);
  vector<i64> ret_pre(m, 0);
  i64 i{0};
  i64 ret{0};
  x %= m;
  while(0 < n){
    --n;
    i_pre[x] = i++;
    ret_pre[x] = ret;
    ret += x;
    x = (x*x)%m;
    if(0 <= i_pre[x]){
      i64 w{i - i_pre[x]};
      ret += (ret - ret_pre[x])*(n/w);
      n %= w;
    }
  }
  cout << ret << endl;
}

int main() {
  cin.tie(nullptr);
  run();
  return 0;
}