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 int n, x, m;
	scanf("%lld %lld %lld", &n, &x, &m);
	long long int i, j;
	long long int v[100005];
	for (i = 0; i < m; i++)
		v[i] = -1;
	j = x;
	for (i = 1;; i++)
	{
		j %= m;
		if (v[j] > 0)
			break;
		v[j] = i;
		j *= j;
	}
	long long int ans = 0;
	if (i >= n)
	{
		j = x % m;
		for (i = 0; i < n; i++)
		{
			ans += j;
			j = j * j % m;
		}
		printf("%lld\n", ans);
		return 0;
	}
	i = x % m;
	while (i != j)
	{
		ans += i;
		i = i * i % m;
		n--;
		if (n == 0)
			break;
	}
	if (n == 0)
	{
		printf("%lld\n", ans);
		return 0;
	}
	n--;
	ans += i;
	if (n == 0)
	{
		printf("%lld\n", ans);
		return 0;
	}
	long long int c = 0, d = 0;
	for (i = j;;)
	{
		c++;
		d += i;
		i = i * i % m;
		if (i == j)
			break;
	}
	ans += n / c * d;
	n %= c;
	i = j;
	for (; n > 0; n--)
	{
		i = i * i % m;
		ans += i;
	}
	printf("%lld\n", ans);
	return 0;
}