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>
#pragma GCC optimize("Ofast")
using namespace std;
long long m, x, p[22], a[22], b[22], len, res = 1;
map<long long, int> Map[22];
inline void init(long long n) {
  for (register long long i = 2; i * i <= n; i++)
    if (n % i == 0) {
      len++;
      p[len] = i;
      while (n % i == 0) n /= i, a[len]++;
    }
  if (n > 1) len++, p[len] = n, a[len] = 1;
}
inline void init2(long long n, int tp) {
  for (register long long i = 2; i * i <= n; i++) {
    while (n % i == 0) Map[tp][i]++, n /= i;
  }
  if (n > 1) Map[tp][n]++;
}
inline long long mul(long long a, long long n, long long mo) {
  long long res = 0;
  n %= mo;
  a %= mo;
  if (mo <= 2000000000) return a * n % mo;
  while (n) {
    if (n & 1) res = res + a < mo ? res + a : res + a;
    a = a + a < mo ? a + a : a + a - mo;
    n >>= 1;
  }
  return res % mo;
}
inline long long power(long long a, long long n, long long mo) {
  long long res = 1;
  if (mo == 1) return 0;
  n %= mo - 1;
  while (n) {
    if (n & 1) res = mul(res, a, mo);
    a = mul(a, a, mo);
    n >>= 1;
  }
  return res;
}
inline void dfs(int dep, long long d) {
  if (dep > len) {
    if (d == 1) return;
    map<long long, int> fac;
    for (register int i = 1; i <= (len); i++)
      if (b[i]) {
        if (b[i] - 1) fac[p[i]] += b[i] - 1;
        for (map<long long, int>::iterator it = Map[i].begin();
             it != Map[i].end(); it++)
          fac[(*it).first] += (*it).second;
      }
    static long long c[22];
    int tot = 0;
    long long phi = 1;
    for (map<long long, int>::iterator it = fac.begin(); it != fac.end();
         it++) {
      tot++, c[tot] = (*it).first;
      for (register int i = 1; i <= ((*it).second); i++) phi *= c[tot];
    }
    long long k = phi;
    for (register int i = 1; i <= (tot); i++) {
      while (k % c[i] == 0 && power(x, k / c[i], d) == 1) k /= c[i];
    }
    res += phi / k;
    return;
  }
  for (register int i = 0; i <= (a[dep]); i++)
    b[dep] = i, dfs(dep + 1, d), d *= p[dep], b[dep] = 0;
}
int main() {
  scanf("%lld%lld", &m, &x);
  init(m);
  for (register int i = 1; i <= (len); i++) init2(p[i] - 1, i);
  dfs(1, 1);
  cout << res;
  return 0;
}