Problem: The campus has m rooms numbered from 0 to m - 1. Also the x-mouse lives in the campus. The x-mouse is not just a mouse: each second x-mouse moves from room i to the room i ⋅ x mod{m} (in fact, it teleports from one room to another since it doesn't visit any intermediate room). Starting position of the x-mouse is unknown.

You are responsible to catch the x-mouse in the campus, so you are guessing about minimum possible number of traps (one trap in one room) you need to place. You are sure that if the x-mouse enters a trapped room, it immediately gets caught.

And the only observation you made is GCD (x, m) = 1.

Input

The only line contains two integers m and x (2 ≤ m ≤ 10^{14}, 1 ≤ x < m, GCD (x, m) = 1) — the number of rooms and the parameter of x-mouse. 

Output

Print the only integer — minimum number of traps you need to install to catch the x-mouse.

Examples

Input

4 3


Output

3


Input

5 2


Output

2

Note

In the first example you can, for example, put traps in rooms 0, 2, 3. If the x-mouse starts in one of this rooms it will be caught immediately. If x-mouse starts in the 1-st rooms then it will move to the room 3, where it will be caught.

In the second example you can put one trap in room 0 and one trap in any other room since x-mouse will visit all rooms 1..m-1 if it will start in any of these rooms.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using std::cin;
using std::cout;
const int maxn = 200200;
const int mod = 998244353;
std::vector<long long> d, prm;
long long m, x;
inline void factor(long long x) {
  for (long long i = 1; i * i <= x; ++i) {
    if (x % i == 0) {
      d.push_back(i);
      if (i * i != x) d.push_back(x / i);
    }
  }
  sort(d.begin(), d.end());
}
inline void fctprm(long long x) {
  for (long long i = 2; i * i <= x; ++i)
    if (x % i == 0) {
      prm.push_back(i);
      do x /= i;
      while (x % i == 0);
    }
  if (x != 1) prm.push_back(x);
}
inline long long phi(long long x) {
  long long ret = x;
  for (long long i : prm) {
    if (x % i == 0) {
      ret /= i;
      ret *= i - 1;
    }
  }
  return ret;
}
inline long long pow(long long a, long long b, long long mod) {
  long long ans = 1;
  for (; b; b >>= 1, a = (__int128)a * a % mod)
    if (b & 1) {
      ans = (__int128)ans * a % mod;
    }
  return ans;
}
inline long long ln(long long x, long long mod, long long phimod) {
  long long ret = phimod;
  for (long long i : prm)
    while (ret % i == 0 && pow(x, ret / i, mod) == 1) ret /= i;
  return ret;
}
int main() {
  std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> m >> x;
  factor(m);
  fctprm(m);
  fctprm(phi(m));
  sort(prm.begin(), prm.end());
  prm.erase(unique(prm.begin(), prm.end()), prm.end());
  long long ans = 1;
  for (long long gcd : d)
    if (gcd != m) {
      long long mod = m / gcd, cnt = phi(mod);
      ans += cnt / ln(x % mod, mod, cnt);
    }
  cout << ans << '\n';
}