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 <atcoder/all>
#include <bits/stdc++.h>
using namespace std;
//using namespace atcoder;
#define int long long
   template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
    const int MAX =3001;
const int MOD =998244353;


  
signed main(){
  int n,x,m;
  cin>>n>>x>>m;
  int k=1,s[m];fill(s,s+m,MOD*MOD);
  while(s[x]==MOD*MOD){
    s[x]=k;
    k++;
    x=x*x%m;
  }
  int ans=0;
  
  for(int i=0;i<m;i++)if(s[i]<=n)ans+=i*(1+(s[x]<=s[i])*(n-s[i])/(k-s[x]));
cout<<ans<<endl;

}
