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<vector>
using LL = long long;
using namespace std;

int main() {
	LL N, X, M;
	cin >> N >> X >> M;
	LL an = X, sum = 0;
	vector<LL> Sum;
	Sum.push_back(0);
	vector<LL> exsists(100010, -1);
	for(int i = 1; i<= N; i++){
		sum += an;
		Sum.push_back(sum);
		if (exsists[an] == -1) {
			exsists[an] = i;
		}
		else {
			LL l = exsists[an];
			LL r = i;
			LL Sdiff = Sum[r] - Sum[l];
			LL diff = r - l;

			LL shuki = (N - r) / diff;
			LL amari = (N - r) % diff;
			sum = sum + Sdiff * shuki + Sum[l + amari] - Sum[l];
			break;
		}
		//cout << i << "\t" << an << "\t" << sum << endl;
		an = an * an % M;
	}

	cout << sum << endl;


	return 0;
}
