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 <stdio.h>
int main(){
  long long int n,a[3],ans=0,temp;
  int m,plus_minus=1;
  scanf("%lld%lld%d",&n,&a[0],&m);
  ans+=a[0];
  for(long long int i=1;i<n;i++){
      a[1+plus_minus]=(a[1-plus_minus]*a[1-plus_minus])%m;
      ans+=a[1+plus_minus];
      if(a[1+plus_minus]<=1){
        temp=i+1;
        break;
      }
      plus_minus*=-1;
  }
  if(a[1+plus_minus]==1){
      ans+=n-temp;
  }
  printf("%lld",ans);
   return 0;
}