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;
long long multi(long long a, long long b, long long c) {
  return ((a * b - (long long)(a / (long double)c * b + 1e-5) * c + c) % c +
          c) %
         c;
}
inline long long power(long long a, long long b, long long m,
                       long long ans = 1ll) {
  for (; b; b >>= 1, a = multi(a, a, m))
    if (b & 1ll) ans = multi(ans, a, m);
  return ans;
}
const int NN = 10000020;
int flag[NN];
long long prime[666555];
int pcnt;
void prepare() {
  for (int i = 2; i < NN; i++) {
    if (!flag[i]) {
      prime[pcnt++] = i;
      if ((long long)i * i > NN) continue;
      for (int j = i * i; j < NN; j += i) {
        if (!flag[j]) flag[j] |= 1;
      }
    }
  }
}
long long fac[111];
int sz;
inline long long get_phi(long long x) {
  long long tmp = x;
  sz = 0;
  for (int i = 0; ((prime[i]) * (prime[i])) <= tmp; i++) {
    if (tmp % prime[i] == 0ll) {
      fac[sz++] = prime[i];
      x -= x / prime[i];
      while (!(tmp % prime[i])) tmp /= prime[i];
    }
  }
  if (tmp > 1ll) {
    fac[sz++] = tmp;
    x -= x / tmp;
  }
  return x;
}
long long x;
inline long long calc(long long n) {
  if (n == 1ll) return 1ll;
  long long p = x % n;
  long long phi = get_phi(n);
  long long tmp = phi;
  get_phi(phi);
  for (int i = 0; i < sz; i++) {
    while (phi % fac[i] == 0ll && power(p, phi / fac[i], n) == 1ll) {
      phi /= fac[i];
    }
  }
  return tmp / phi;
}
int main() {
  prepare();
  long long m;
  long long ans = 0;
  cin >> m >> x;
  long long d = 1ll;
  for (; d * d < m; d++) {
    if (m % d == 0) {
      ans += calc(d);
      ans += calc(m / d);
    }
  }
  if (d * d == m) {
    ans += calc(d);
  }
  cout << ans << endl;
  return 0;
}