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;
typedef std::pair<int,int> pair2;
typedef std::pair<int,pair2> pair3;
const ll INF = 1LL<<60;
const  ll NUM=10e8+7;


int main(){
  ll N,X,M;
  cin>>N>>X>>M;
  
  ll ans=0;
  vector<ll> dist(M,-1);//カウント
 ll count;
  count=1;
while(1){
  //cout<<X;
  dist[X]=count;
  ans=ans+X;
  //cout<<"  "<<ans<<endl;
  if(count==N){cout<<ans<<endl;return 0;}
  X=(X*X)%M;
  //cout<<"  "<<X<<endl;
  count++;
  if(dist[X]!=-1){break;}
  }
//cout<<ans<<endl;
  ll cicle=count-dist[X];
  //cout<<cicle<<"."<<X<<endl;
  for(ll i=0;i<cicle;i++){
    ll num=(N-dist[X])/cicle;
    ans=ans+X*num;
    //cout<<X<<" "<<X*num<<endl;
    X=(X*X)%M;
  }
  cout<<ans<<endl;
}
