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>
//#include <atcoder/all>
using namespace std;
//using namespace atcoder;
#define ll long long
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rep1(i, n) FOR(i, 1, n+1)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; --i)
#define whole(x) (x).begin(),(x).end()
#define rwhole(x) (x).rbegin(), (x).rend()
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define P pair<ll, ll>
#define debug(var) cerr << "[" << #var << "] " << var << endl
template<typename T1, typename T2>
bool chmax(T1 &a,T2 b){if(a<b){a=b;return true;}else return false;}
template<typename T1, typename T2>
bool chmin(T1 &a,T2 b){if(a>b){a=b;return true;}else return false;}
#define vi vector<int>
#define vl vector<ll>
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define pr(s) cout << (s) << '\n'
const ll mod = 1000000007;
const int dx[] = {-1,0,1,0};
const int dy[] = {0,-1,0,1};
const int INF = 1001001001;
const ll INFll = 1E+18;


int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    ll n, x, m;
    cin >> n >> x >> m;
    ll ans = 0;
    vector<P> memo(m, P(-1, -1));
    for (ll i=0; i<n; i++) {
        if (memo[x].second>0) {
            ll loop = i - memo[x].first;
            //debug(loop);
            ll sum_loop = ans - memo[x].second + x;
            ll num = (n-i-1)/loop;
            ans += sum_loop * num;
            i += loop*num;
        }
        else {
            memo[x] = {i, ans+x};
        }
        ans += x;
        x = (x*x)%m;
        if (x==0) return -1;
    }
    
    pr(ans);
    return 0;
}