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: Accepted)

#include <stdio.h>

int main()
{
	long long N;
	int X, M;
	scanf("%lld %d %d", &N, &X, &M);
	
	int j, flag[100001] = {};
	long long i, A[100001], sum[100001] = {0, X};
	for (i = 1, A[1] = X, flag[X] = 1; i < N && flag[A[i] * A[i] % M] == 0; i++) {
		A[i+1] = A[i] * A[i] % M;
		flag[A[i+1]] = i + 1;
		sum[i+1] = sum[i] + A[i+1];
	}
	if (i >= N) printf("%lld\n", sum[i]);
	else {
		j = flag[A[i] * A[i] % M];
		printf("%lld\n", (sum[i] - sum[j-1]) * ((N - j + 1) / (i + 1 - j)) + sum[(N-j+1)%(i+1-j)+j-1]);
	}
	fflush(stdout);
	return 0;
}