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: Runtime Error)

#include <bits/stdc++.h>
#define rep(i,N) for(int i = 0 ; i < int(N);i++)
#define all(x) (x.begin(),x.end())
using namespace std ;
typedef long long ll;
typedef pair <int , int >pi;
typedef vector <int> vi ;
typedef vector <string> si;



int main (){
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	int b,c;
	long long int a,sum = 0;
	cin >>a >> b >> c;
	
	vector<int> v(a);
	for (int i = 0 ;i < a ;i++){
		if(b < c){
			v[i]=b;
			// cout << b << endl;
			b *= b;
			b %= c;
		}
	}
	for (int i = 0 ;i < a ;i++){
		sum += v[i];
	}
	cout << sum ;


	return 0 ;
}