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;

int main() {
	long long n,x,m,sum=0;
	cin >> n >> x >> m;
	long long a=x,b=0,totyu;
	for(long long i=0;i<n;i++){
		sum+=a;
		a=(a*a)%m;
		if(a==0){
			break;
		}else if(a==1){
			sum+=n-i-1;
			break;
		}else if(a==b){
			sum=totyu+(sum-totyu)*((n-101)/(i-100));
			i=100+(i-100)*((n-101)/(i-100));
		}
		if(i==100){
			b=a;
			totyu=sum;
		}
	}
	cout << sum << endl;
}