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 <iomanip>
#include <algorithm>
#include <assert.h>
#include <complex>
#include <utility>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#include <tuple>
#include <cmath>
#include <bitset>
#include <cctype>
#include <set>
#include <map>
#include <unordered_map>
#include <numeric>
#include <functional>
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(ll i=ll(a);i<ll(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define rrep(i,a) for(ll i=ll(a-1);i>=0;--i)
#define all(x) (x).begin(),(x).end()
#define PRINT(V) cout << V << "\n"
#define SORT(V) sort((V).begin(),(V).end())
#define RSORT(V) sort((V).rbegin(), (V).rend())
using namespace std;
using ll = long long;
using ull = unsigned long long;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
inline void Yes(bool condition){ if(condition) PRINT("Yes"); else PRINT("No"); }
template<class itr> void cins(itr first,itr last){
    for (auto i = first;i != last;i++){
        cin >> (*i);
    }
}
template<class itr> void array_output(itr start,itr goal){
    string ans = "",k = " ";
    for (auto i = start;i != goal;i++) ans += to_string(*i)+k;
    if (!ans.empty()) ans.pop_back();
    PRINT(ans);
}
ll gcd(ll a, ll b) {
    return a ? gcd(b%a,a) : b;
}

const ll INF = 1e18;
const ll MOD = 1000000007;
const ll MOD2 = 998244353;
const ll MOD3 = 1e6;
const ll EPS = 1e-10;
int sgn(const double a){
    return (a < -EPS ? -1 : (a > EPS ? +1 : 0));
}
typedef pair<ll,ll> P;
typedef tuple<ll,ll,ll> tri;
typedef pair<long double,long double> point;
typedef complex<long double> Point;
const ll MAX = 3000;
constexpr ll nx[4] = {-1,0,1,0};
constexpr ll ny[4] = {0,1,0,-1};
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n,x,m;
    cin >> n >> x >> m;
    vector<ll> done(100005,0);
    ll a = x,cnt = 0;
    done[a] = 0;
    while (true){
        a = a*a%m;
        if (done[a]){
            cnt++;
            break;
        }
        cnt++;
        done[a] = cnt;
    }
    ll sum = a,ans = x;
    ll s = a;
    while (true){
        s = s*s%m;
        if (s == a) break;
        sum += s;
    }
    ll len = cnt-done[a];
    if (n <= done[a]){
        rep(i,n-1){
            x = x*x%m;
            ans += x;
        }
        PRINT(ans);
        return 0;
    }
    else{
        rep(i,done[a]-1){
            x = x*x%m;
            ans += x;
        }
        n -= done[a];
        ans += sum*(n/len);
        n %= len;
        rep(i,n){
            ans += a;
            a = a*a%m;
        }
        PRINT(ans);
    }
}