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;
typedef long long ll;
typedef unsigned long long ull;

int main() {
	ll N, X, M;
	cin >> N >> X >> M;

	vector<ll> A(M,-1); // 回数
	vector<ll> B(M,-1); // Cumm
	ll c = 0;
	int f = 0;
	for(ll i = 1; i <= N; i++){
		c = c + X;
		if(A[X] == -1 || f == 1){ // まだ来てない
			A[X] = i;
			B[X] = c;
		}else{
			ll j = A[X]; // 周期 = i - j
			ll d = c - B[X];
			ll r = (N - i) / (i - j); // 周回数
			c += r * d;
			i = i + r * (i - j); // 周期回数
			f = 1;
		}
		X = (X * X) % M;
	}

	cout << c << endl;
	return(0);
}
