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>
#define MAX 10000000000
int main(void)
{
    int i;
    int n, m, x;
    int F[MAX];
    int sum = 0;

    scanf("%d", &n);
    scanf("%d", &x);
    scanf("%d", &m);

    F[0] = x;
    sum = F[0];

    for(i = 0; i < n-1; i++){
        F[i + 1] = (F[i] * F[i]) % m;
        sum += F[i + 1];
    }

    printf("%d\n", sum);

    return 0;
}