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>
int m[10000001], pr[664580], v;
long long M, X;
long long phi(long long n) {
  long long r = n;
  for (int i = 1; i <= v && (long long)pr[i] * pr[i] <= n; i++)
    if (n % pr[i] == 0) {
      r = r / pr[i] * (pr[i] - 1);
      do n /= pr[i];
      while (n % pr[i] == 0);
    }
  if (n > 1) r = r / n * (n - 1);
  return r;
}
long long MUL(long long a, long long b, long long c) {
  long long tmp = a * b - (long long)((long double)a * b / c) * c;
  return (tmp % c + c) % c;
}
long long POWER(long long a, long long b, long long c) {
  long long r = 1;
  for (; b; b >>= 1) {
    if (b & 1) r = MUL(r, a, c);
    a = MUL(a, a, c);
  }
  return r;
}
long long ANS(long long M, long long X) {
  if (M == 1) return 1;
  long long d = phi(M), _d = d;
  std::vector<long long> fac_d;
  for (int i = 1; i <= v && (long long)pr[i] * pr[i] <= _d; i++)
    if (_d % pr[i] == 0) {
      fac_d.push_back(pr[i]);
      do _d /= pr[i];
      while (_d % pr[i] == 0);
    }
  if (_d > 1) fac_d.push_back(_d);
  _d = 1;
  for (long long e : fac_d)
    while (d % e == 0 && POWER(X, d / e, M) == 1) {
      d /= e;
      _d *= e;
    }
  return _d;
}
int main() {
  for (int i = 2; i <= 10000000; i++) {
    if (!m[i]) m[i] = pr[++v] = i;
    for (int j = 1; j <= v && pr[j] * i <= 10000000 && pr[j] <= m[i]; j++)
      m[pr[j] * i] = pr[j];
  }
  scanf("%lld%lld", &M, &X);
  long long O = 0;
  for (long long i = 1; i * i <= M; i++)
    if (M % i == 0) {
      O += ANS(i, X % i);
      if (i * i < M) O += ANS(M / i, X % (M / i));
    }
  printf("%lld\n", O);
  return 0;
}