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() {
  long n,x,m,ans=0;
  bool aaa=false;
  cin >> n >> x >> m;
  vector<long> a(5000000),aruisekiwa(5000001),syutugen(m,-1);
  syutugen[x]=0;
  a[0]=x;
  aruisekiwa[1]=x;
  ans+=a[0],n--;
  for(long i=1;n!=0;i++) {
    a[i]=a[i-1]*a[i-1]%m;
    aruisekiwa[i+1]=aruisekiwa[i]+a[i];
    if(syutugen[a[i]]==-1||aaa) {
      syutugen[a[i]]=i;
      ans+=a[i],n--;
    } else {
      aaa=true;
      ans+=n/(i-syutugen[a[i]])*(aruisekiwa[i]-aruisekiwa[syutugen[a[i]]]);
      //n=n%(i-syutugen[a[i]]);
      ans+=(aruisekiwa[syutugen[a[i]]+n%(i-syutugen[a[i]])]-aruisekiwa[syutugen[a[i]]]);
      //i=syutugen[a[i]]-1;
      break;
    }
  }
  cout << ans << endl;
}