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;
const int maxn =3e5+13;

int a[maxn];
int vis[maxn];
ll b[maxn];
int main()
{
	ll n;
	int x,m;
	scanf("%lld%d%d",&n,&x,&m);
	a[1]=x;
	int cnt =1;
	vis[a[1]]=1;
	int last,fir;
	while(1)
	{
		a[cnt+1]=1ll*a[cnt]*a[cnt]%m;
	    cnt++;
	    if(!vis[a[cnt]]) vis[a[cnt]]=cnt;
	    else 
		{
		last = cnt - vis[a[cnt]];
		fir = vis[a[cnt]];
		break;}
	}
	for(int i=1;i<=cnt;i++)
	{
		b[i]=b[i-1]+1ll*a[i];
	}
	
	ll ans=b[fir-1]+(b[cnt-1]-b[fir-1])*((n-fir+1)/(last))+b[(n-fir+1)%last+fir-1]-b[fir-1];
	printf("%lld",ans);
 } 