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: Wrong Answer)

#define LOCAL

#include <bits/stdc++.h>

using namespace std;

#define IOS ios::sync_with_stdio(false);cin.tie(0)
#define all(x) x.begin(), x.end()
#define ff first
#define ss second
#define LLINF 0x3f3f3f3f3f3f3f3f
#define INF 0x3f3f3f3f


#define LOCAL

// Copied from Gennady-Korotkevich's template

template <typename A, typename B>
string to_string(pair<A, B> p);

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);

string to_string(const string& s) {
	return '"' + s + '"';
}

string to_string(const char* s) {
	return to_string((string)s);
}

string to_string(bool b) {
	return (b ? "true" : "false");
}

string to_string(vector<bool> v) {
	bool first = true;
	string res = "{";
	for (int i = 0; i < static_cast<int>(v.size()); i++) {
		if (!first) {
			res += ", ";
		}
		first = false;
		res += to_string(v[i]);
	}
	res += "}\n";
	return res;
}

template <size_t N>
string to_string(bitset<N> v) {
	string res = "";
	for (size_t i = 0; i < N; i++) {
		res += static_cast<char>('0' + v[i]);
	}
	return res;
}

template <typename A>
string to_string(A v) {
	bool first = true;
	string res = "{";
	for (const auto& x : v) {
		if (!first) {
			res += ", ";
		}
		first = false;
		res += to_string(x);
	}
	res += "}\n";
	return res;
}

template <typename A, typename B>
string to_string(pair<A, B> p) {
	return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
	return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ")";
}

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
	return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}

template <typename A, typename B, typename C, typename D, typename E>
string to_string(tuple<A, B, C, D, E> p) {
	return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + "," + to_string(get<4>(p)) + ")";
}

void debug_out() {
	cerr << endl;
}

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
	cerr << " " << to_string(H);
	debug_out(T...);
}

#ifdef LOCAL
#define debug(...) cerr << "\n[" << #__VA_ARGS__ << "]:\n", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

// End of Gennady-Korotkevich's template 

using llong = long long;
using PII = pair<int, int>;

const llong MOD = 998244353LL;
const double PI = acos(-1);

int main() {
	IOS;
	llong n, x, m; cin >> n >> x >> m;
	if (n == 1) {
		cout << x;
		return 0;
	}
	vector<int> pos(m, -1);
	pos[x] = 0;
	llong cur = x;
	llong curIdx = 0;
	int startCycle, endCycle;
	vector<llong> ansans;
	ansans.push_back(cur);
	while (1) {
		cur *= cur;
		cur %= m;
		curIdx++;
		if (pos[cur] != -1) {
			startCycle = pos[cur];
			endCycle = curIdx - 1;
			break;
		}
		pos[cur] = curIdx;
		ansans.push_back(cur + ansans.back());
	}
	llong cycleLength = endCycle - startCycle + 1;
	if (n <= startCycle) {
		cout << ansans[n - 1];
		return 0;
	}
	llong ans; 
	if (startCycle != 0) ans = ansans[startCycle - 1];
	else ans = 0;
	n -= startCycle; 
	llong sumcycle;
	if (startCycle == 0) sumcycle = ansans[endCycle];
	else sumcycle = ansans[endCycle] - ansans[startCycle - 1];
	ans += (n / cycleLength) * sumcycle;
	n %= cycleLength;
	ans += ansans[startCycle + n -1] - ansans[startCycle - 1];
	cout << ans;
}