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 <map>
#include <tuple>

using namespace std;

int main()
{
	long long n, x, m;
	cin >> n >> x >> m;
	long long ans = 0;
	map<long long, tuple<long long, long long, int>> mp; //xが出てきた数, sum, i
	bool f = true;
	for (long long i = 1; i <= n; i++) {
		ans += x;
		if (f) {
			if (get<0>(mp[x]) == 1) {
				f = false;
				
				long long t = ans - get<1>(mp[x]);
				ans += (n - i) / (i - get<2>(mp[x])) * t;
				n -= (n - i) / (i - get<2>(mp[x])) * (i - get<2>(mp[x]));
			}
			else {
				get<0>(mp[x])++;
				get<1>(mp[x]) = ans;
				get<2>(mp[x]) = i;
			}
		}
		x = x * x % m;
	}
	cout << ans << endl;

	return 0;
}