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() {
int64_t n,x,m,a,so,sy,ma;
  a=0;
  so=0;
  sy=0;
  ma=0;
  cin>>n>>x>>m;
  vector<int64_t> vec(m+1);
  vec.at(0)=x%m;
  for(int64_t i=0;i<m;i++){
  vec.at(i+1)=(vec.at(i)*vec.at(i))%m;
  }
  for(int64_t i=m;i>=0;i--){
  if(vec.at(m)==vec.at(i)){a=i;}
  }
  for(int64_t i=a;i<=(n-a-1)%(m-a)+a;i++){
  so=so+vec.at(i);}
  for(int64_t i=a;i<m;i++){
    sy=sy+vec.at(i);
  }
  for(int64_t i=0;i<=a-1;i++){
   ma=ma+vec.at(i);
  }
  cout<<ma+so+((n-a-1)/(m-a))*sy<<endl;
} 