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>
using namespace std;

int main()
{
    long long int N, X, M;
    cin >> N >> X >> M;

    long long int A[N];

    A[0] = X;
    for (int i = 1; i < N; i++)
    {
        A[i] = (A[i - 1] * A[i - 1]) % M;
    }
    long long int sum = 0;

    for (int i = 0; i < N; i++)
    {
        sum += A[i];
    }
    cout << sum << endl;

    return 0;
}