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;

int main(){
  long long N,X,M;
  cin >> N >> X >> M;
  vector<long long> V(M,-1);
  
  long long sum = 0;
  for(int i=0;i<N;i++){
    if(V[X]==-1){
      V[X] = i;
      sum += X;
      X *= X;
      X %= M;
    }
    else{
      long long k = i - V[X], Ws = 0;
      vector<long long> W(k);
      for(long long j=0;j<M;j++){
        if(V[X]<=V[j] && V[j]<i){
          W[V[j]-V[X]] = j;
          Ws += j;
        }
      }
      
      sum += Ws * ((N-i)/k);
      for(int j=0;j<(N-i)%k;j++){
        sum += W[j];
      }
      break;
    }
  }
  
  cout << sum << endl;
  return 0;
}
