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;
int f[10000005], k;
long long m, x, i, j, prime[1000005], ans;
inline long long qmult(long long x, long long y, long long p) {
  long long z = (long double)x / p * y;
  long long res = (unsigned long long)x * y - (unsigned long long)z * p;
  return (res + p) % p;
}
long long qpow(long long a, long long b, long long mod) {
  long long s = a % mod, ans = 1;
  while (b) {
    if (b & 1) ans = qmult(ans, s, mod);
    s = qmult(s, s, mod);
    b /= 2;
  }
  return ans;
}
long long phi(long long n) {
  int i;
  long long s = n;
  for (i = 1; i <= k && prime[i] * prime[i] <= n; i++)
    if (n % prime[i] == 0) {
      s = s / prime[i] * (prime[i] - 1);
      while (n % prime[i] == 0) n /= prime[i];
    }
  if (n != 1) s = s / n * (n - 1);
  return s;
}
void Add(long long n) {
  long long p = phi(m / n);
  long long tmp = p, i, s = p;
  for (i = 1; i <= k && prime[i] * prime[i] <= tmp; i++)
    if (tmp % prime[i] == 0) {
      while (s % prime[i] == 0 && qpow(x, s / prime[i], m / n) == 1)
        s /= prime[i];
      while (tmp % prime[i] == 0) tmp /= prime[i];
    }
  if (tmp != 1)
    if (qpow(x, s / tmp, m / n) == 1) s /= tmp;
  ans += p / s;
}
int main() {
  for (i = 2; i <= 10000000; i++) {
    if (f[i] == 0) prime[++k] = i;
    for (j = 1; j <= k && prime[j] * i <= 10000000; j++) {
      f[prime[j] * i] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  cin >> m >> x;
  for (i = 1; i * i < m; i++)
    if (m % i == 0) {
      Add(i);
      Add(m / i);
    }
  if (i * i == m) Add(i);
  cout << ans << endl;
}