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::vector;
long long mul(long long a, long long b, long long c) {
  long long r = a * b - (long long)((long double)a * b / c + 0.5) * c;
  return r < 0 ? r + c : r;
}
long long fpow(long long a, long long b, long long c) {
  long long d = 1, x = a;
  while (b) {
    if (b & 1) d = mul(d, x, c);
    x = mul(x, x, c);
    b >>= 1;
  }
  return d;
}
struct node {
  long long p;
  int k;
};
long long n, K, ans;
vector<node> div(long long n) {
  vector<node> d;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) {
      int j = 0;
      while (n % i == 0) ++j, n /= i;
      d.push_back((node){i, j});
    }
  if (n > 1) d.push_back((node){n, 1});
  return d;
}
void calc(long long d) {
  long long m = n / d;
  auto fe = div(m);
  long long phi = m;
  for (auto f : fe) phi = phi / f.p * (f.p - 1);
  fe = div(phi);
  long long jie = phi, k = K % m;
  for (auto f : fe)
    for (int j = 0; j < f.k; j++)
      if (fpow(k, jie / f.p, m) == 1)
        jie /= f.p;
      else
        break;
  ans += phi / jie;
}
int main() {
  scanf("%lld%lld", &n, &K);
  for (long long i = 1; i * i <= n; i++)
    if (n % i == 0) {
      calc(i);
      if (i * i < n) calc(n / i);
    }
  printf("%lld", ans);
  return 0;
}