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;
long long m, x;
int cnt1, cnt2;
long long prime1[10001000], prime2[10000100];
int tot;
long long d[10001000];
long long ksc(long long x, long long y, long long mod) {
  long long res = 0;
  res = (x * y - (long long)((double)x * y / mod + 0.1) * mod) % mod;
  if (res < 0) res += mod;
  return res;
}
long long qpow(long long y, long long n, long long mod) {
  long long res = 1;
  long long now = y % mod;
  while (n) {
    if (n & 1) res = ksc(res, now, mod);
    now = ksc(now, now, mod);
    n /= 2;
  }
  return res;
}
long long Euler(long long y) {
  for (int i = 0; i < cnt1; i++) {
    if (y % prime1[i] == 0) {
      y = y / prime1[i] * (prime1[i] - 1ll);
    }
  }
  return y;
}
int main() {
  scanf("%lld%lld", &m, &x);
  for (long long i = 1; i * i <= m; i++) {
    if (m % i == 0) {
      d[tot++] = i;
      if (i * i != m) d[tot++] = m / i;
    }
  }
  long long m1 = m;
  for (long long i = 2; i * i <= m1; i++) {
    if (m1 % i == 0) {
      prime1[cnt1++] = i;
      while (m1 % i == 0) m1 /= i;
    }
  }
  if (m1 > 1) prime1[cnt1++] = m1;
  m1 = Euler(m);
  for (long long i = 2; i * i <= m1; i++) {
    if (m1 % i == 0) {
      prime2[cnt2++] = i;
      while (m1 % i == 0) m1 /= i;
    }
  }
  if (m1 > 1) prime2[cnt2++] = m1;
  if (cnt2 == 0) {
    prime2[cnt2++] = 1;
  }
  long long res = 0;
  for (int i = 0; i < tot; i++) {
    long long f = Euler(d[i]);
    long long l = f;
    if (l > 1) {
      for (int j = 0; j < cnt2; j++) {
        while (l % prime2[j] == 0 && qpow(x, l / prime2[j], d[i]) == 1) {
          l /= prime2[j];
        }
      }
    }
    res += f / l;
  }
  printf("%lld\n", res);
  return 0;
}