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>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vll = vector<long long>;
using vvi = vector<vector<int>>;
using pii = pair<int,int>;
using pll = pair<long,long>;
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define pb push_back

vll rem(1e5+5,0),vec(1e5+5,0);

int main() {
  ll n,x,m,first=0,period=0;
  cin>>n>>x>>m;
  ll sum=0,ans=0;

  for(ll i=1;i<=m-1;i++) {
    rem[i]=x;
    if(vec[x]!=0) {
      first=vec[x];
      period=i-first;
    }
    vec[x]=i;
    x*=x;
    x%=m;
  }
  if(first!=0) {
    for(ll i=first;i<=first+period-1;i++) sum+=rem[i];
  } else {
    for(ll i=1;i<=m-1;i++) sum+=rem[i];
    ans+=sum*(n/period);
    n%=period;
    for(ll i=1;i<=n;i++) ans+=rem[i];
    cout<<ans<<endl;
    return 0;
  }

  if(n<=first) {
    for(ll i=1;i<=n;i++) ans+=rem[i];
  } else {
    for(ll i=1;i<=first-1;i++) ans+=rem[i];
    n-=first-1;
    ans+=sum*(n/period);
    n%=period;
    for(ll i=first;i<=first+n-1;i++) ans+=rem[i];
  }

  cout<<ans<<endl;
}