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>
#define mod 998244353
#define maxn 200005
using namespace std;
long long n,x,m,tt=99999999999ll;
int vis[maxn];
long long arr[maxn];
int main()
{
	long long sum=0;
	cin>>n>>x>>m;
	for (int i=1;i<=n;i++)
	{
		if (vis[x]!=0)
		{
			arr[i]=arr[i-1]+x;
			long long zhouqihe,zhouqi;
			zhouqihe=arr[i]-arr[vis[x]];
			zhouqi=i-vis[x];
			long long lef=n-i+1;
			long long tim=lef/zhouqi;
			sum+=zhouqihe*tim;
			tt=i+tim*zhouqi;
			break;
		}
		sum+=x;
		arr[i]=arr[i-1]+x;
		vis[x]=i;
		x=x*x%m;
	}
	for (tt;tt<=n;tt++)
	{
		sum+=x;
		x=x*x%m;
	}
	cout<<sum<<endl;
}