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;

long long n, x, m;
vector<vector<long long>> memo, sum;

int main() {
  cin >> n >> x >> m;
  memo.assign(60, vector<long long>(m, 0));
  sum.assign(60, vector<long long>(m, 0));
  for (int i = 0; i < m; ++i) memo[0][i] = (long long)i * i % m;
  for (int i = 0; i < m; ++i) sum[0][i] = i;
  for (int i = 1; i < 60; ++i)
    for (int j = 0; j < m; ++j) {
      long long mid = memo[i - 1][j];
      memo[i][j] = memo[i - 1][mid];
      sum[i][j] = sum[i - 1][j] + sum[i - 1][mid];
    }
  long long res = 0;
  for (int i = 0; i < 60; ++i)
    if (n >> i & 1) {
      res += sum[i][x];
      x = memo[i][x];
    }
  cout << res << endl;
  return 0;
}