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;
#define ll long long
#define all(V) V.begin(),V.end()
const ll MOD = 1000000007;

ll T[100010], S[100010];

int main() {
	for (int i = 0;i < 100010;i++)T[i] = -1, S[i] = 0;
	ll N, X, M, fx, ans = 0;
	cin >> N >> X >> M;
	N--;
	T[X] = N;
	S[X] = X;
	ans += X;
	N--;
	while (N >= 0) {
		X = X * X % M;
		ans += X;
		if (T[X] >= 0) {
			ll tr = T[X] - N;
			ll ntr = N / tr;
			ans += ntr * (ans - S[X]);
			N %= tr;
			N--;
			break;
		}
		T[X] = N;
		S[X] = ans;
		N--;
	}
	while (N >= 0) {
		X = X * X % M;
		ans += X;
		N--;
	}
	cout << ans << endl;
}
