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 m, x, a[100];
long long mul(long long a, long long b, long long mod) {
  long long ret = 0;
  while (b) {
    if (b % 10000) ret = (ret + (b % 10000) * a) % mod;
    a = (a * 10000) % mod;
    b /= 10000;
  }
  return ret;
}
long long phi(long long n) {
  long long ret = n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ret -= ret / i;
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) ret -= ret / n;
  return ret;
}
long long quick(long long a, long long b, long long c) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = mul(ret, a, c);
    a = mul(a, a, c);
    b >>= 1;
  }
  return ret;
}
long long count(long long k) {
  long long n = phi(k);
  for (int i = 1; i <= a[0]; i++)
    if (n % a[i] == 0)
      while (n % a[i] == 0) {
        if (quick(x, n / a[i], k) == 1)
          n /= a[i];
        else
          break;
      }
  return n;
}
long long calc(long long d) { return phi(m / d) / count(m / d); }
int main() {
  cin >> m >> x;
  long long M = phi(m);
  for (long long i = 2; i * i <= M; i++)
    if (M % i == 0) {
      a[++a[0]] = i;
      while (M % i == 0) M /= i;
    }
  if (M > 1) a[++a[0]] = M;
  long long ans = 0;
  for (long long d = 1; d * d <= m; d++)
    if (m % d == 0) {
      ans += calc(d);
      if (d * d < m) ans += calc(m / d);
    }
  cout << ans;
  return 0;
}