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 <iostream>
#include<string>
#include <algorithm>
#include <math.h>
using namespace std;
 
int main() {
  long int N, X, M, a[100002]={}, ans = 0, ak, anss[100002]={}, ansnum[100002]={};
  cin >> N >> X >> M;
  for(long int i = 0; i<M; i++){
    a[i] = i*i%M;
  }
  ak = X;
  for(long int j = 0; j<N; j++){
    ans += ak;
    if(anss[ak]==0){
      anss[ak]= ans;
      ansnum[ak] = j;
    }
    else{
      long int num = (N-j-1)/(j-ansnum[ak]);
      ans += num*(ans-anss[ak]);
      j+=num*(j-ansnum[ak]);
    }
    ak = a[ak];
    if(ak == 0)break;
  }
  cout << ans << endl;
  return 0;
}