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 namespace std;
const long long N = 2e5 + 1;
const long long inf = LLONG_MAX;
vector<long long> dvs, pri;
long long phi[N];
long long ord[N];
long long x, p;
long long mul(long long a, long long b, long long mod) {
  if (a < b) swap(a, b);
  long long ans = 0;
  string s = "";
  while (b) {
    s += (b % 10 + '0');
    b /= 10;
  }
  for (long long i = s.size() - 1; i >= 0; --i)
    ans = (ans * 10 + (s[i] - '0') * a) % mod;
  return ans;
}
long long gcd(long long a, long long b) { return a == 0 ? b : gcd(b % a, a); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
long long qpow(long long a, long long exp, long long mod) {
  long long ans = 1;
  while (exp) {
    if (exp & 1) ans = mul(ans, a, mod);
    a = mul(a, a, mod);
    exp >>= 1;
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> p >> x;
  for (long long i = 1; i * i <= p; ++i) {
    if (p % i) continue;
    dvs.push_back(i);
    if (i < p / i) dvs.push_back(p / i);
  }
  sort(dvs.begin(), dvs.end());
  p = dvs.back();
  for (long long i = 2; i * i <= p; ++i) {
    if (p % i) continue;
    pri.push_back(i);
    while (!(p % i)) p /= i;
  }
  if (p > 1) pri.push_back(p);
  for (long long i = 1; i < dvs.size(); ++i)
    for (long long p : pri) {
      if (dvs[i] % p) continue;
      if (dvs[i] == p) {
        phi[i] = p - 1;
        for (long long j = 1; j * j < p; ++j) {
          if ((p - 1) % j) continue;
          if (qpow(x, j, dvs[i]) == 1) {
            ord[i] = j;
            break;
          }
          if (qpow(x, (p - 1) / j, dvs[i]) == 1) ord[i] = (p - 1) / j;
        }
        break;
      }
      long long X = dvs[i], Y = p;
      phi[i] = p - 1;
      X /= p;
      while (!(X % p)) {
        phi[i] *= p;
        X /= p;
        Y *= p;
      }
      if (X == 1) {
        phi[i] =
            phi[lower_bound(dvs.begin(), dvs.end(), Y / p) - dvs.begin()] * p;
        ord[i] = ord[lower_bound(dvs.begin(), dvs.end(), Y / p) - dvs.begin()];
        if (qpow(x, ord[i], dvs[i]) != 1) ord[i] *= p;
      } else {
        phi[i] = phi[lower_bound(dvs.begin(), dvs.end(), Y) - dvs.begin()] *
                 phi[lower_bound(dvs.begin(), dvs.end(), X) - dvs.begin()];
        ord[i] = lcm(ord[lower_bound(dvs.begin(), dvs.end(), Y) - dvs.begin()],
                     ord[lower_bound(dvs.begin(), dvs.end(), X) - dvs.begin()]);
      }
      break;
    }
  long long ans = 1;
  for (long long i = 1; i < dvs.size(); ++i) ans += phi[i] / ord[i];
  cout << ans << endl;
}