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>
using namespace std;
long N,X,M;
int vis[1<<17];
long pre[1<<17];
int main()
{
	cin>>N>>X>>M;
	for(int i=0;i<M;i++)vis[i]=-1;
	long ans=0;
	int tu=0;
	while(N>0)
	{
		if(vis[X]>=0)
		{
			int be=tu-vis[X];
			ans+=(ans-pre[X])*(N/be);
			N%=be;
		}
		if(N>0)
		{
			vis[X]=tu++;
			pre[X]=ans;
			ans+=X;
			N--;
			X=X*X%M;
		}
	}
	cout<<ans<<endl;
}
