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;
long long m, x, t, tt;
const int N = 200000;
vector<long long> pd;
map<long long, long long> g;
long long v[N], vv[N];
long long phi(long long x) {
  long long up = sqrt(x) + 1;
  long long n = x;
  for (int i = 2; i <= up && i <= x; i++)
    if (x % i == 0) {
      pd.push_back(i);
      x /= i;
      while (x % i == 0) x /= i;
      n = n / i * (i - 1);
    }
  if (x > 1) {
    pd.push_back(x);
    n = n / x * (x - 1);
  }
  return n;
}
long long phi2(long long x) {
  for (int i = 0; i < pd.size(); i++)
    if (x % pd[i] == 0) x = x / pd[i] * (pd[i] - 1);
  return x;
}
void div(long long x, long long* v, long long& t) {
  long long up = sqrt(x) + 1;
  t = 0;
  for (int i = 1; i <= up; i++)
    if (x % i == 0) {
      if (i > x / i) break;
      if (i <= x / i) v[t++] = i;
      if (i < x / i) v[t++] = x / i;
    }
}
long long mul(long long a, long long b) {
  long long ret = 0;
  while (b) {
    ret = (ret + a * (b & ((1 << 16) - 1))) % m;
    b >>= 16;
    a = (a << 16) % m;
  }
  return ret;
}
long long qp(long long x, long long y, long long m) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = mul(ret, x) % m;
    x = mul(x, x) % m;
    y >>= 1;
  }
  return ret;
}
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
int main() {
  scanf("%lld%lld", &m, &x);
  long long p = phi(m);
  int l = 0;
  div(p, v, t);
  div(m, vv, tt);
  sort(vv, vv + tt);
  for (int i = 0; i < t; i++) {
    long long xk = qp(x, v[i], m);
    long long c = gcd(m, xk ? xk - 1 : m - 1);
    if (!g.count(c) || v[i] < g[c]) g[c] = v[i];
  }
  for (int j = tt - 1; j >= 0; j--) {
    long long d = vv[j];
    long long c = g[d];
    if (!c) continue;
    for (int i = 0; i < pd.size(); i++)
      if ((d % pd[i] == 0) && (!g[d / pd[i]] || c < g[d / pd[i]]))
        g[d / pd[i]] = c;
  }
  long long ans = 0;
  for (int i = 0; i < tt; i++) ans += phi2(vv[i]) / g[vv[i]];
  printf("%lld\n", ans);
  return 0;
}
