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;
  cin >> N >> X >> M;
  
  long ans = 0;
  vector<pair<long, long>> pre(M);

  for (long i = 0; i < N; i++) {
    auto [p_i, p_ans] = pre.at(X);
    if (!p_i) {
      pre.at(X) = {i, ans};
    } else {
      long len = i - p_i;
      long cnt = (N - i - 1) / len;
      ans += (ans - p_ans) * cnt;
      i += len * cnt;
    }
    ans += X;
    X = X * X % M;
  }
  cout << ans << "\n";
}