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;
vector<long long> pd;
long long phi(long long a) {
  if (a == 2) {
    pd.push_back(2);
    return 1;
  }
  long long up = sqrt(a) + 1;
  long long ans = a;
  for (long long i = 2; i <= up; i++) {
    if (a % i == 0) {
      ans = (i - 1) * (ans / i);
      pd.push_back(i);
    }
    while (a % i == 0) a /= i;
  }
  if (a > 1) {
    pd.push_back(a);
    ans = (a - 1) * (ans / a);
  }
  return ans;
}
long long phi2(long long a) {
  for (auto it = pd.begin(); it != pd.end(); it++) {
    if (a % *it == 0) a = (*it - 1) * (a / *it);
  }
  return a;
}
void div(long long n, vector<long long>& v) {
  long long up = sqrt(n) + 1;
  for (long long i = 1; i <= up; i++) {
    if (n % i == 0) {
      if (i <= n / i) v.push_back(i);
      if (i < n / i) v.push_back(n / i);
    }
  }
}
long long mul(long long x, long long y, long long mod) {
  long long ans = 0;
  while (y) {
    ans = (ans + x * (y & ((1 << 8) - 1))) % mod;
    y >>= 8;
    x = (x << 8) % mod;
  }
  return ans;
}
long long modPow(long long b, long long e, long long mod) {
  if (e == 0) return 1;
  return e % 2 ? mul(b, (modPow(mul(b, b, mod), e / 2, mod)) % mod, mod)
               : modPow(mul(b, b, mod), e / 2, mod) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long m, x;
  cin >> m >> x;
  vector<long long> divM;
  div(m, divM);
  long long ph = phi(m);
  vector<long long> divPh;
  div(ph, divPh);
  sort(divPh.begin(), divPh.end());
  vector<long long> xk;
  for (auto it = divPh.begin(); it != divPh.end(); it++) {
    xk.push_back(modPow(x, *it, m));
  }
  long long ans = 0;
  for (auto itD = divM.begin(); itD != divM.end(); itD++) {
    for (int i = 0; i < int(divPh.size()); i++) {
      if ((xk[i] - 1) % *itD == 0) {
        ans += phi2(*itD) / (divPh[i]);
        break;
      }
    }
  }
  cout << ans << '\n';
  return 0;
}