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;
#define rep(i, n) for(int i = 0;i < (int)(n);i++)

int main() {
  long long x, m, ls = 0, k = 0, j = 1, s, rest, n, S = 0;
  bool c = false;
  cin >> n >> x >> m;
  vector<int> A(m+5), f(m, 1);
  A[0] = x;
  x = x*x % m;
  while(f[x]) {
    A[j] = x;
    f[x] = 0;
    x = x*x % m;
    j++;
  }
  rep(i, j) {
    if(A[i] == x) {
      c = true;
      s = i;
    }
    if(c) {
      ls += A[i];
      k++;
    }
  }
  if(n <= s) {
    rep(i, n) S += A[i];
  } else {
    rep(i, s) S += A[i];
    S += ls * ((n-s) / k);
    rest = n - s - ((n-s) / k) * k;
    rep(i, rest) S += A[s+i];
  }
  cout << S << endl;
}