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){
    int n, x, m;

    long int  result = 0;

    int *a, *kako;

    int i, j;

    int maeno = -1;
    int nidome = -1;

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

    a = (int*)malloc(sizeof(int) * (m));
    kako = (int*)malloc(sizeof(int) * (m)); //来てたらそのindex


    for(i = 0; i < m; i++){
        kako[i] = -1;
        a[i] = 0;
    }
    a[1] = x;



    for(i = 2; i <= n; i++){
//        printf("ok3 ");
//        printf("");

        long int kari = (long int)a[i - 1] * (long int)a[i - 1];
        kari %= (long int) m;
        a[i] = (int) kari;
//        printf("%d\n", a[i]);
//        printf("%d\n", i);


        if(kako[a[i]] == -1)
            kako[a[i]] = i;
        else{
            maeno = kako[a[i]];
            nidome = i;
            break;
        }

    }


    if(maeno == -1){
        for(i = 1; i <= n; i++)
            result += (long int) a[i];
    }
    else{
//        int roop = 0;
        // ループ回数 (n - maeno) / (nidome - maeno)
        // 余り (n - maeno) % (nidome - maeno)

        for(i = maeno; i < nidome; i++){
            result += (long int) a[i];
        }

        result *= (long int)((n - maeno) / (nidome - maeno));

        for(i = 0; i < maeno; i++){
            result += (long int) a[i];
        }


        for(i = maeno; i < (n - maeno) % (nidome - maeno); i++){
            result += (long int) a[i];
        }

    }
    printf("%ld\n", result);


    free(a);
    free(kako);

    return 0;
}