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<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
const int maxm=1e5;
ll n;
int x,m,s,id;
ll a[maxm+5],pres[maxm+5],ss[maxm+5],tots;
int used[maxm+5];
int main(){
	scanf("%lld",&n);
	scanf("%d %d",&x,&m);
	a[1]=x;
	for(s=1;!used[a[s]];s++){
		used[a[s]]=s;
		a[s+1]=a[s]*a[s]%m;
		pres[s]=pres[s-1]+a[s];
	}
	tots+=pres[used[a[s]]-1];
	for(int i=used[a[s]];i<s;i++)
		ss[i-used[a[s]]+1]=ss[i-used[a[s]]]+a[i];
	n=n-used[a[s]]+1; s-=used[a[s]];
	//printf("%lld %d\n",n,s);
	if(a[s]==0) printf("%lld\n",pres[s-1]);
	else printf("%lld\n",ss[s]*(n/s)+tots+ss[n%s]);
	return 0;
}
