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<iostream>
#include<vector>
using namespace std;
int main() {
  long long N;
  int X,M;
  cin >> N >> X >> M;
  vector<long long> p(M,-1),A(M+3);
  A[0] = X;
  int i = 0,j;
  bool f = true;
  while(f) {
    i++;
    A[i] = (A[i-1]*A[i-1])%M;
    if(p[A[i]] > 0)f = false;
    else p[A[i]] = i;
  }
  long long ans = X;
  if(p[A[i]] < N)for(j=0;j<M;j++) {
    if(p[j] >= p[A[i]] && p[j] < i)ans += (N-p[A[i]])/(i-p[A[i]])*j;
    if(p[j] >= p[A[i]] && p[j] <= p[A[i]] + (N-p[A[i]])%(i-p[A[i]]) - 1)ans += j;
  }
  for(j=0;j<M;j++)if(p[j] > 0 && p[j] < min(p[A[i]],N))ans += j;
  cout << ans << endl;
}