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<bits/stdc++.h>
using namespace std;
using ll = long long;

int main() {
    ll n, x, m;
    cin >> n >> x >> m;
    set<ll> s;
    vector<ll> v;
    v.push_back(x);
    s.insert(x);
    ll lsz;
    vector<ll>::iterator it;
    while (true) {
        ll a = (v.back() * v.back()) % m;
        ll sz = s.size();
        s.insert(a);
        if (sz == s.size()) {
            it = find(v.begin(), v.end(), a);
            lsz = s.size() - distance(v.begin(), it);
            break;
        }
        v.push_back(a);
    }
    ll idx = distance(v.begin(), it);
    ll bsum = 0;
    if (it != v.begin()) {
        bsum = accumulate(v.begin(), it, 0ll);
    }
    ll lsum = accumulate(it, v.end(), 0ll);
    ll ans = 0;
    if (v.size() >= n) {
        cout << accumulate(v.begin(), v.begin() + n, 0ll) << endl;
    }
    else {
        ans += bsum;
        n -= distance(v.begin(), it);
        ans += (n / lsz) * lsum;
        ans += accumulate(it, it + (n % lsz), 0ll);
        cout << ans << endl;
    }
}