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;
  cin>>n>>x>>m;
  if(n<m){
    int64_t a=0;
    for(int64_t i=x,j=0;j<n;i=i*i%m,j++)
      a+=i;
    cout<<a<<endl;
    return 0;
  }
  vector<int64_t> c(m,0);
  vector<int64_t> o(m,m);
  int l=m;
  int p=m;
  for(int64_t i=x,j=0;;i=i*i%m,j++){
    if(c.at(i)!=0){
      l=i;
      p=j-o.at(i);
      break;
    }
    c.at(i)++;
    o.at(i)=j;
  }
  for(int64_t i=l,j=0;j<p;i=i*i%m,j++){
    c.at(i)=0;
    c.at(i)+=(n-o.at(l))/p+(j<(n-o.at(l))%p?1:0);
  }
  int64_t s=0;
  for(int i=0;i<m;i++)
    s+=c.at(i)*i;
  cout<<s<<endl;
}