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>
typedef long long ll;
using namespace std;
int main(void){
  ll n,x,m; cin>>n>>x>>m; --n;
  vector<bool> c(m,false);
  ll res=x;
  while(n-->0){
    x=x*x%m;
    if(c[x]==false) res+=x,c[x]=true;
    else{
      ll cnt=0,cnt2=0;
      vector<bool> d(m,false);
      while(d[x]==false){
        cnt+=x;
        d[x]=true;
        x=x*x%m;
        ++cnt2;
      }
      res+=cnt*(n/cnt2);
      n-=n/cnt2*cnt2;
      while(n-->=0) res+=x,x=x*x%m;
    }
  }
  cout<<res;
  return 0;
}
