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;
using ll=long long;
ll N,X,M;
int main(){
  cin>>N>>X>>M;
  if(N<=M){
  ll ans=0;ll a=X;
    for(ll i=0;i<N;i++){
    ans+=a;
      a=(a*a)%M;
    }cout<<ans<<endl;
      
  }else{
  set<ll>S;ll a=X;ll I,r;
    for(ll i=0;i<N;i++){
    if(S.count(a)){I=i-1;r=a;break;}
      S.insert(a);
      a=(a*a)%M;
    }a=X;
    ll ans=0;ll J;
    for(ll i=0;i<N;i++){
    if(a!=r){ans+=a;a=(a*a)%M;}else{J=i;break;}
     }
  ll A=(I-J+1);ll B=(N-J)%A;ll C=(N-J)/A;
    ll sum=0;ll R=r;
    for(ll i=J;i<=I;i++){
    sum+=R;R=(R*R)%M;
    }ll y=0;ll R1=r;
    for(ll i=J;i<J+B;i++){
    y+=R1;R1=(R1*R1)%M;
    }ans+=(sum*C+y);
      
      cout<<ans<<endl;
   }
  return 0;
}
  