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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
 
int main() {
  long long n,x,m,ans,loopans,ac,jdg;cin >> n >> x >> m;
  ans=x;ac=x;loopans=0;jdg=0;
  vector<long long> a(m,0),loopcount(3,0);
  
  for (long long i=1;i<n;i++) {
    ac=ac*ac%m;
	if (a[ac]==0) {a[ac]=i;ans+=ac;}
	else {
	  loopcount[1]=i;
	  loopcount[2]=i-a[ac];
	  break;
	}
  }
  if (loopcount[1]==0) cout << ans << endl;
  else{
  
  for (long long i=0;i<loopcount[2];i++) {
    if (loopcount[1]+i>n-1) {
	  jdg=1;
	  break;
	}
	loopans+=ac;
	ac=ac*ac%m;
  }
  if (jdg==1) cout << ans+loopans << endl;
  else {
    ans+=((int((n-loopcount[1])/loopcount[2]))*(loopans));
	loopans=0;
	for (long long i=0;i<(n-loopcount[1])%loopcount[2];i++) {     
	  loopans+=ac;
	  ac=ac*ac%m;
    }
	cout << ans+loopans << endl;
  } 
  }
}