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;
int n, tot, pri[9] = {2, 3, 5, 7, 11, 13, 17, 19, 23}, top;
long long m, x, a[100005], b[100005], f[100005], l[100005], prime[10005];
map<long long, int> id;
long long gcd(long long x, long long y) {
  if (!y)
    return x;
  else
    return gcd(y, x % y);
}
long long mul(long long x, long long y, long long mo) {
  x %= mo;
  y %= mo;
  long long ans = (x * y - (long long)((double)x * y / mo + 0.1) * mo) % mo;
  ans += ans < 0 ? mo : 0;
  return ans;
}
long long ksm(long long x, long long y, long long mo) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = mul(ans, x, mo);
    x = mul(x, x, mo);
    y >>= 1;
  }
  return ans;
}
bool MR(long long n) {
  if (n == 2) return 1;
  if (n % 2 == 0) return 0;
  long long w = n - 1;
  int lg = 0;
  while (w % 2 == 0) w /= 2, lg++;
  for (int i = 0; i < 9; i++) {
    if (n == pri[i]) return 1;
    long long x = ksm(pri[i], w, n);
    for (int j = 0; j < lg; j++) {
      long long y = mul(x, x, n);
      if (x != 1 && x != n - 1 && y == 1) return 0;
      x = y;
    }
    if (x != 1) return 0;
  }
  return 1;
}
long long rho(long long n) {
  long long c = rand() * rand() % (n - 1) + 1, x1 = rand() * rand() % n,
            x2 = x1, k = 1, p = 1;
  for (int i = 1; p == 1; i++) {
    x1 = (mul(x1, x1, n) + c) % n;
    if (x1 == x2) return 1;
    p = gcd(n, abs(x1 - x2));
    if (i == k) k <<= 1, x2 = x1;
  }
  return p;
}
void divi(long long n) {
  if (n == 1) return;
  if (MR(n)) {
    b[++tot] = n;
    return;
  }
  long long p = 1;
  while (p == 1) p = rho(n);
  divi(p);
  divi(n / p);
}
void pre() {
  for (long long i = 1; i * i <= m; i++)
    if (m % i == 0) {
      a[++n] = i;
      if (m / i != i) a[++n] = m / i;
    }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) id[a[i]] = i;
  long long tmp = m;
  for (long long i = 2; i * i <= tmp; i++)
    if (tmp % i == 0) {
      prime[++top] = i;
      while (tmp % i == 0) tmp /= i;
    }
  if (tmp > 1) prime[++top] = tmp;
}
int main() {
  scanf("%lld%lld", &m, &x);
  pre();
  for (int i = 1; i <= n; i++) f[i] = m / a[i] - 1;
  for (int j = 1; j <= top; j++)
    for (int i = 1; i <= n; i++)
      if (a[i] % prime[j] == 0) f[id[a[i] / prime[j]]] -= f[i];
  for (int i = n; i >= 1; i--) {
    l[i] = a[i];
    for (int j = 1; j <= top; j++)
      if (l[i] % prime[j] == 0) l[i] = l[i] / prime[j] * (prime[j] - 1);
    tot = 0;
    divi(l[i]);
    sort(b + 1, b + tot + 1);
    tot = unique(b + 1, b + tot + 1) - b - 1;
    for (int j = 1; j <= tot; j++)
      while (l[i] % b[j] == 0 && ksm(x, l[i] / b[j], a[i]) == 1) l[i] /= b[j];
  }
  long long ans = 1;
  for (int i = 1; i < n; i++) {
    int j;
    for (j = 1; a[j] != m / a[i]; j++)
      ;
    ans += f[i] / l[j];
  }
  printf("%lld\n", ans);
  return 0;
}