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: Wrong Answer)

#include "bits/stdc++.h"
using namespace std;

int main() {
	long long N, X, M;
	cin >> N >> X >> M;
	set<long long>s;
	map<long long, long long>p;
	vector<long long>Ans(M + 10);
	s.insert(X);
	p[X] = 0;
	long long ans = X;
	Ans[0] = 0;
	Ans[1] = ans;
	for (long long i = 1; i < N; ++i) {
		X *= X;
		X %= M;
		ans += X;
		Ans[i + 1] = ans;
		if (s.count(X)) {
			ans = 0;
			long long memo = N - i;
			memo %= (i - p[X]);
			ans += Ans[p[X]];
			ans += (Ans[i - 1] - Ans[p[X] - 1])*((N - p[X]) / (i - p[X]));
			ans += (Ans[p[X] + memo - 1] - Ans[p[X] - 1]);
			cout << ans << endl;
			return 0;
		}
		s.insert(X);
		p[X] = i;
	}
	cout << ans << endl;
	return 0;
}