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 <iostream>
#include <set>
using namespace std;
typedef long long ll;

int main() {
	ll n, x; int m; cin >> n >> x >> m;
	set<ll> s; ll res = 0;
	ll cur = x; while (!s.count(cur)) {
		s.insert(cur); res += cur;
		cur = (cur * cur) % m;
	}
	set<ll> t; ll cycle = 0;
	while (!t.count(cur)) {
		t.insert(cur); cycle += cur;
		cur = (cur * cur) % m;
	}

	if (n <= (int)s.size()) {
		res = 0; while (n--) { res += x; x = (x * x) % m; }
		cout << res << '\n'; return 0;
	}
	n -= (int)s.size(); res += (n / t.size())*cycle;
	x = cur; n %= t.size();
	while (n--) { res += x; x = (x * x) % m; }
	cout << res << '\n';
	cin.ignore(2); return 0;
}