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: Accepted)

#include <iostream>
#include <cstdint>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cctype>
#include <cassert>
#include <climits>
#include <string>
#include <bitset>
#include <cfloat>
#include <unordered_set>
#pragma GCC optimize("Ofast")
using namespace std;
typedef long double ld;
typedef long long int ll;
typedef unsigned long long int ull;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef vector<string> vs;
typedef vector<ll> vll;
typedef vector<pair<int, int> > vpii;
typedef vector<vector<int> > vvi;
typedef vector<vector<char> > vvc;
typedef vector<vector<string> > vvs;
typedef vector<vector<ll> > vvll;

#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rrep(i, n) for (int i = 1; i <= (n); ++i)
#define irep(it, stl) for (auto it = stl.begin(); it != stl.end(); it++)
#define drep(i, n) for (int i = (n)-1; i >= 0; --i)
#define mes(a) cout << (a) << endl
#define dmes(a, b) cout << (a) << " " << (b) << endl
#define re0 return 0
#define mp(p, q) make_pair(p, q)
#define pb(n) push_back(n)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define Sort(a) sort(a.begin(), a.end())
#define rSort(a) sort(a.rbegin(), a.rend())
#define MATHPI acos(-1)
#define itn int;
int dx[8] = { 1, 0, -1, 0, 1, -1, -1, 1 };
int dy[8] = { 0, 1, 0, -1, 1, 1, -1, -1 };
template <class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template <class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
struct io { io() { ios::sync_with_stdio(false); cin.tie(0); } };
const int INF = INT_MAX;
const ll LLINF = 1LL << 60;
const ll MOD = 1000000007;
const double EPS = 1e-9;

signed main(void)
{
    ll n, x, m;
    cin >> n >> x >> m;
    vb used(m*2);
    vll one_cycle;
 
    ll start = 0;
    ll now = x;
    
 	rep(i, n) {
        used[now] = true;
        one_cycle.pb(now);
        now = now * now % m;
        if (used[now]) {
            for(ll j = 0; j < (ll)one_cycle.size(); ++j) if (one_cycle[j] == now) {
                start = j;
                break;
            }
        break;
        }
    }
 
    ll s1 = 0, s2 = 0, loop = one_cycle.size() - start;
    
    rep(i, start) s1 += one_cycle[i];
    
    for(int i = start; i < one_cycle.size(); ++i) s2 += one_cycle[i];
    
    ll ans = s1 + (n - start) / loop * s2;
    
    for(int i = 0; i < (n - start) % loop; ++i) {
        ans += now;
        now = now * now % m;
    }
    
    mes(ans);
    
    return 0;
}