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 rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<ll,ll>P;

const int MOD=1000000007;
const int INF=0x3f3f3f3f;
const ll INFL=0x3f3f3f3f3f3f3f3f;

int main(){
	ll n,x,m;cin>>n>>x>>m;
	map<ll,int>mp;
	map<ll,ll>mp_sum;
	ll a=x;
	int cnt=0;
	ll sum=0;
	while(1){
		cnt++;
		mp[a]=cnt;
		mp_sum[a]=sum;
		sum+=a;
		if(cnt==n){
			cout<<sum<<endl;
			return 0;
		}
		a=(a*a)%m;
		if(mp.count(a)){
			int s=(cnt+1)-mp[a];
			ll ans=mp_sum[a];
			n-=mp[a]-1;
			ll d=sum-mp_sum[a];
			ans+=(n/s)*d;
			n%=s;
			rep(i,n){
				ans+=a;
				a=(a*a)%m;
			}
			cout<<ans<<endl;
			return 0;
		}
	}
}