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<stdio.h>
int main(){
  long long n,m,x,ans=0,s[100010],f[100010]={},lc,ls,nokori,sum=0,i,a;
  scanf("%lld %lld %lld",&n,&x,&m);
  int d[10]={1,2,3,4,2,6,7,1};
  f[x]=1;
  s[x]=x;
  sum=x;
  /*if(n==1){
    printf("%lld\n",x);
    return 0;
    }//*/
  for(i=1;i<n;i++){
    a=(x*x)%m;
    //a=d[x];
    sum=s[x]+a;
    if(f[a])break;
    f[a]=i;
    s[a]=sum;
    x=a;
  }
  ls=sum-s[a];
  lc=i-f[a];
  nokori=n-i-1;
  if(lc==0){
    printf("%lld\n",x*n);
    return 0;
  }
  //printf("%lld\n",i);
  //printf("%lld %lld %lld %lld\n",ls,lc,sum,nokori);
  if(nokori<0)nokori=0;
  ans=sum+(nokori/lc)*ls;
  nokori=nokori%lc;
  x=a;
  for(i=0;i<nokori;i++){
    a=(x*x)%m;
    //a=d[x];
    ans+=a;
    x=a;
  }
  printf("%lld\n",ans);
  return 0;
}