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;
const int N = 10000010;
vector<long long> get_divs(long long n) {
  vector<long long> ret;
  ret.push_back(n);
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      if (n / i != i) {
        ret.push_back(n / i);
      }
    }
  }
  sort(ret.begin(), ret.end());
  return ret;
}
vector<long long> get_primes(long long n) {
  vector<long long> ret;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) ret.push_back(n);
  return ret;
}
long long euler(long long n, vector<long long> &p) {
  long long ret = n;
  for (long long x : p) {
    if (ret % x == 0) {
      ret = ret / x * (x - 1);
    }
  }
  return ret;
}
long long mul(long long x, long long y, long long mod) {
  long long res = (x * y - (long long)((double)x * y / mod + 0.1) * mod) % mod;
  if (res < 0) res += mod;
  return res;
}
long long pow_mod(long long a, long long b, long long md) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = mul(ret, a, md);
    b >>= 1;
    a = mul(a, a, md);
  }
  return ret;
}
void solve(long long m, long long x) {
  vector<long long> divs = get_divs(m);
  vector<long long> pris_m = get_primes(m);
  long long eul = euler(m, pris_m);
  vector<long long> pris_phim = get_primes(eul);
  long long ret = 1;
  for (long long d : divs) {
    long long f = euler(d, pris_m);
    long long k = f;
    for (long long p : pris_phim) {
      while (k % p == 0 && pow_mod(x % d, k / p, d) == 1) {
        k /= p;
      }
    }
    ret += f / k;
  }
  cout << ret << endl;
}
int main() {
  long long m, x;
  cin >> m >> x;
  solve(m, x);
  return 0;
}