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}.

Examples

Input

6 2 1001


Output

1369


Input

1000 2 16


Output

6


Input

10000000000 10 99959


Output

492443256176507
#include <bits/stdc++.h>
using namespace std;
using ll = long long;

int main() {
	ll n, x, m; cin >> n >> x >> m;
	vector<ll> cnt;
	ll rem = x;
	ll tmp = m;
	while (tmp--) {
		cnt.push_back(rem);
		rem = rem * rem % m;
		if (find(cnt.begin(), cnt.end(), rem) != cnt.end())
			break ;
	}
	ll mae = distance(cnt.begin(), find(cnt.begin(), cnt.end(), rem));
	ll maeSUM = accumulate(cnt.begin(), cnt.begin() + mae, 0LL);
	ll loop = cnt.size() - mae;
	ll loopSUM = accumulate(cnt.begin() + mae, cnt.end(), 0LL);
	ll res = maeSUM;
	n -= mae;
	ll div = n / loop;
	ll mod = n % loop;
	res += loopSUM * div;
	res += accumulate(cnt.begin() + mae, cnt.begin() + mae + mod, 0LL);
	cout << res << '\n';
    return 0;
}
