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;
long long n;
long long x,m;
int sa[100100];
long long add[100010];
int main()
{
	cin>>n>>x>>m;
	int Bit=1;
	long long l,r;
	for(long long a=x;;a=(a*a)%m,Bit++)
	{
		if(!sa[a]) sa[a]=Bit;
		else
		{
			l=sa[a],r=Bit-1;
			break;
		}
		add[Bit]=add[Bit-1]+a;
	}
	long long len=r-l+1;
	long long ans=add[min(n,l-1)];
	n=max(0ll,n-l+1);
	if(n) ans+=(add[r]-add[l-1])*(n/len)+(add[n%len+l-1]-add[l-1]);
	cout<<ans;
	return 0;
}