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)
#define repit(it, li) for(auto it=li.begin(); it!=li.end(); it++)
using namespace std;
using ll = long long;
using P = pair<int,int>;

// static int mi[100000];
// static ll ms[100000];

int main(){
	ll n;
	int x, m;
	cin>>n>>x>>m;

	vector<ll> mi(m, -1);
	vector<ll> ms(m);
	// rep(i, m) mi[i]= -1;

	ms[x]=x;
	mi[x]=1;
	int c=x;
	ll s=x;
	bool skipped=false;
	for(ll i=2; i<=n; i++){
		c=(ll)c*c%m;
		s+=c;
		if(mi[c]!= -1){
			if(!skipped){
				ll len=i-mi[c];
				ll ds= s-ms[c];
				ll b=(n-i)/len;
				s+= ds*b;
				i+= b*len;
				skipped=true;
			}
		}
		else{
			mi[c]=i;
			ms[c]=s;
		}
	}

	cout<<s<<endl;
	return 0;
}
