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 <iostream>
#include <algorithm>
#include <bits/stdc++.h>
#define FAST ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define scanArray(a,n) for(int i = 0; i < n; i++){cin >> a[i];}

using namespace std;

int main()
{
    long long n;
    long long x,m,sum=0;
    cin>>n>>x>>m;

    sum+=x;
    for (long long  i=1;i<n;i++){
        x=(x*x)%m;
        sum=sum+x;
    }
    cout<<sum;
    return 0;
}