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 N=1e5+5,M=2e4+5,inf=0x3f3f3f3f,mod=1e9+7;
#define mst(a,b) memset(a,b,sizeof a)
#define lx x<<1
#define rx x<<1|1
#define reg register
#define PII pair<int,int>
#define fi first
#define se second
#define pb push_back
#define il inline
ll n,x,m;
int vis[N],id=1;
ll pre[N];
int main(){
	scanf("%lld%lld%lld",&n,&x,&m);
	ll ans=0;
	while(n--){
		vis[x]=id++;
		pre[x]=ans;
		ans+=x;
		x=x*x%m;
		if(vis[x]>0){
			int jie=(id-vis[x]);
			ans+=(ans-pre[x])*(n/jie);
			n%=jie; 
		}
	} 
	printf("%lld\n",ans);
	return 0;
}