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;
inline long long read() {
  char ch = getchar();
  long long s = 0, w = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * w;
}
long long m, x;
vector<long long> d, val;
long long prime[1000010], tot;
int vis[1000010 * 10];
inline long long PHI(long long x) {
  long long tmp = x, ans = x;
  for (register long long i = 1; i <= tot; i++) {
    if (prime[i] > sqrt(x)) break;
    if (tmp % prime[i] == 0) {
      ans = ans / prime[i] * (prime[i] - 1);
      while (tmp % prime[i] == 0) tmp /= prime[i];
    }
  }
  if (tmp > 1) ans = ans / tmp * (tmp - 1);
  return ans;
}
inline long long mul(long long a, long long b, long long mod) {
  return ((unsigned long long)a * b -
          (unsigned long long)((long double)a / mod * b) * mod + mod) %
         mod;
}
inline long long ksm(long long a, long long b, long long mod) {
  long long ans = 1;
  a %= mod;
  while (b) {
    if (b & 1) ans = mul(ans, a, mod);
    b >>= 1;
    a = mul(a, a, mod);
  }
  return ans;
}
inline long long Solve(long long val) {
  long long t = PHI(val), v = t, d = t;
  for (register long long i = 1; i <= tot; i++) {
    long long p = prime[i];
    if (p > sqrt(t)) break;
    while (v % p == 0 && ksm(x, v / p, val) == 1) v /= p;
    while (d % p == 0) d /= p;
  }
  if (d != 1 && ksm(x, v / d, val) == 1) v /= d;
  return t / v;
}
inline void prepare(long long Maxn) {
  for (register long long i = 2; i <= Maxn; i++) {
    if (!vis[i]) prime[++tot] = i;
    for (register long long j = 1; j <= tot && i * prime[j] <= Maxn; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  prepare(1e7);
  m = read(), x = read();
  long long ans = 0;
  for (register long long i = 1; i <= sqrt(m); i++) {
    if (m % i == 0) {
      ans += Solve(i);
      if (i * i != m) ans += Solve(m / i);
    }
  }
  cout << ans << endl;
  return 0;
}