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;

ll n, ans, r;
int x, m, s, t;
int a[100000];

int main() {
	cin >> n >> x >> m;
	fill(a, a + m, -1);
	r = n;
	for (int i = 0; i < n; ++i) {
		if (a[x] < 0) {
			a[x] = i;
			ans += x;
			x = (ll) x * x % m;
			--r;
		} else {
			s = a[x];
			t = i - a[x];
			break;
		}
	}
	ll c = 0;
	for (int i = 0; i < t; ++i) {
		c += x;
		x = (ll) x * x % m;
	}
	ans += c * (r / t);
	r %= t;
	for (int i = 0; i < r; ++i) {
		ans += x;
		x = (ll) x * x % m;
	}
	cout << ans << endl;
}
