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 <stdio.h>
#include <stdlib.h>

int main (void) {
    long int n, sum = 0;
    int m, x, i;
    scanf("%ld %d %d", &n, &x, &m);
    long int a[4 * m + 2];
    a[0] = x;
    for (i = 1; i < 4 * m + 2; i++) a[i] = (a[i - 1] * a[i - 1]) % m;

    if (m >= n) for (i = 0; i < n; i++) sum += a[i];
    else {
        for (i = 0; i < m; i++) sum += a[i];
        if (a[m - 1] == 1) {
            sum += n - m;
        } else if (a[m - 1] != 0) {
            sum *= (n / m);
            for (i = 0; i < n % m; i++) sum += a[i];
        }
    }
    printf("%ld\n", sum);
    return 0;
}