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;
void get_prim(long long x, long long prim[], int &top) {
  top = 0;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      prim[++top] = i;
      while (x % i == 0) x /= i;
    }
  }
  if (x != 1) prim[++top] = x;
  sort(prim + 1, prim + 1 + top);
}
void get_fac(long long x, long long fac[], int &top) {
  top = 0;
  for (long long i = 1; i * i <= x; i++) {
    if (x % i == 0) {
      fac[++top] = i;
      if (i * i != x) fac[++top] = x / i;
    }
  }
  sort(fac + 1, fac + 1 + top);
}
long long mul(long long x, long long y, long long mod) {
  long long res = (x * y - (long long)((double)x * y / mod + 0.1) * mod) % mod;
  if (res < 0) res += mod;
  return res;
}
long long fsp(long long x, long long y, long long mod) {
  x %= mod;
  long long res = 1;
  while (y) {
    if (y & 1ll) res = mul(res, x, mod);
    x = mul(x, x, mod);
    y >>= 1ll;
  }
  return res;
}
int tot, cnt;
long long prims[100010], fact[100010], facs[100010];
long long Euler(long long x) {
  for (int i = 1; i <= tot; i++)
    if (x % prims[i] == 0) x = x / prims[i] * (prims[i] - 1ll);
  return x;
}
long long m, x;
long long solve() {
  long long eul = Euler(m);
  long long res = 0;
  int tt = 0;
  if (eul > 1)
    get_prim(eul, fact, tt);
  else
    fact[++tt] = 1;
  for (int i = 1; i <= cnt; i++) {
    long long f = Euler(facs[i]);
    long long l = f;
    if (l > 1) {
      for (int j = 1; j <= tt; j++)
        while (l % fact[j] == 0 && fsp(x, l / fact[j], facs[i]) == 1)
          l /= fact[j];
    }
    res = res + f / l;
  }
  return res;
}
int main() {
  scanf("%lld%lld", &m, &x);
  get_fac(m, facs, cnt);
  get_prim(m, prims, tot);
  printf("%lld", solve());
}