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;
signed long long M, X;
vector<signed long long> Ds;
vector<pair<signed long long, int>> Ps;
signed long long totient(signed long long v) {
  for (auto& p : Ps)
    if (v % p.first == 0) v = v / p.first * (p.first - 1);
  return v;
}
signed long long multpow(signed long long a, signed long long b,
                         signed long long m) {
  signed long long r = 0;
  a %= m;
  while (b) {
    if (b & 1) r += a;
    a <<= 1;
    b >>= 1;
    if (a >= m) a -= m;
  }
  return r % m;
}
signed long long modpow(signed long long a, signed long long p,
                        signed long long m) {
  signed long long r = 1;
  while (p) {
    if (p & 1) r = multpow(r, a, m);
    a = multpow(a, a, m);
    p /= 2;
  }
  return r;
}
signed long long ord(signed long long x, signed long long d) {
  if (x == 1) return 1;
  vector<pair<signed long long, int>> P;
  int i;
  signed long long t = totient(d);
  signed long long v = t;
  for (signed long long a = 2; a * a <= v; a++)
    if (v % a == 0) {
      P.push_back({a, 0});
      while (v % a == 0) P.back().second++, v /= a;
    }
  if (v > 1) P.push_back({v, 1});
  signed long long ret = t;
  for (auto& p : P) {
    int i;
    signed long long dv = 1, t2 = t;
    for (i = 0; i < (p.second); i++) {
      t2 /= p.first;
      if (modpow(x, t2, d) != 1) break;
      ret /= p.first;
    }
  }
  return ret;
}
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> M >> X;
  signed long long M2 = M;
  for (signed long long d = 1; d * d <= M; d++) {
    if (M % d == 0) {
      if (d > 1) Ds.push_back(d);
      if (d * d != M) Ds.push_back(M / d);
    }
    if (d > 1 && M2 % d == 0) {
      Ps.push_back({d, 0});
      while (M2 % d == 0) M2 /= d, Ps.back().second++;
    }
  }
  if (M2 > 1) Ps.push_back({M2, 1});
  signed long long ret = 1;
  for (auto& d : Ds) {
    signed long long phi = totient(d);
    signed long long o = ord(X % d, d);
    ret += phi / o;
  }
  cout << ret << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}