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.
#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
long long m, x;
set<long long> s;
vector<long long> u, v;
long long mm(long long a, long long b, long long c) {
  long long ret = 0;
  for (; b; b >>= 1, a = a * 2 % c)
    if (b & 1) ret = (ret + a) % c;
  return ret;
}
long long qm(long long a, long long b, long long c) {
  long long ret = 1;
  for (a %= c; b; b >>= 1, a = mm(a, a, c))
    if (b & 1) ret = mm(ret, a, c);
  return ret;
}
long long ou(long long a) {
  long long b = a, t = a;
  for (long long i = 2; i * i <= t; i++) {
    if (t % i == 0) {
      b = b / i * (i - 1);
      while (t % i == 0) t /= i;
    }
  }
  if (t != 1) b = b / t * (t - 1);
  return b;
}
long long f(long long a) {
  long long b = ou(a);
  for (int i = 0; i < u.size(); i++)
    if (b % u[i] == 0) {
      if (v[i] % a <= 1) return b / u[i];
    }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> m >> x;
  for (long long i = 1; i * i <= m; i++)
    if (m % i == 0) {
      s.insert(i);
      s.insert(m / i);
    }
  for (long long a = ou(m), i = 1; i * i <= a; i++) {
    if (a % i == 0) {
      u.push_back(i);
      u.push_back(a / i);
    }
  }
  sort(u.begin(), u.end());
  unique(u.begin(), u.end());
  v.resize(u.size());
  for (int i = 0; i < u.size(); i++) {
    v[i] = qm(x, u[i], m);
  }
  long long ans = 0;
  for (auto i : s) {
    ans += f(i);
  }
  cout << ans;
  return 0;
}
