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: Time Limit Exceeded)

#include <cstdio>
#include <map>
using namespace std;
long long n,x,m;
long long a[200005];
map<long long,int>ma;
int l,r;
long long sum,asum;
int main(){
	scanf("%lld%lld%lld",&n,&x,&m);
	a[1]=x;
	for(int i=2;;++i){
		a[i]=a[i-1]*a[i-1]%m;
		if(ma.count(a[i])){
			l=ma[a[i]];
			r=i;
			break;
		}
		ma[a[i]]=i;
	}
	if(n<l){
		for(int i=1;i<=n;++i)sum+=a[i];
		printf("%lld\n",sum);
		return 0;
	}
	for(int i=1;i<l;++i){
		sum+=a[i];
	}
	n-=l;
	++n;
	for(int i=l;i<r;++i)asum+=a[i];
	while(n>=r-l){
		sum+=asum;
		n-=r-l;
	}
	for(int i=l;i<l+n;++i)sum+=a[i];
	printf("%lld\n",sum);
	return 0;
}