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 gcd(long long x, long long y) { return y ? gcd(y, x % y) : x; }
long long lcm(long long x, long long y) { return x / gcd(x, y) * y; }
vector<long long> primeFactors;
long long etf(long long x) {
  long long etf = x;
  for (auto i : primeFactors) {
    if (x % i == 0) {
      etf = etf / i * (i - 1);
      do {
        x /= i;
      } while (x % i == 0);
    }
  }
  return etf;
}
long long modex(long long b, long long e, long long m) {
  long long h = 1;
  while (e) {
    if (e & 1) h = (__int128)h * b % m;
    b = (__int128)b * b % m;
    e >>= 1;
  }
  return h;
}
int main() {
  long long m, x;
  scanf("%lld%lld", &m, &x);
  {
    long long M = m;
    for (long long i = 2; i * i <= m; i++) {
      if (m % i == 0) {
        primeFactors.push_back(i);
        do {
          m /= i;
        } while (m % i == 0);
      }
    }
    if (m > 1) {
      primeFactors.push_back(m);
    }
    m = M;
  }
  vector<long long> mfactors;
  for (long long i = 1; i * i <= m; i++) {
    if (m % i == 0) {
      mfactors.push_back(i);
      mfactors.push_back(m / i);
    }
  }
  sort(mfactors.begin(), mfactors.end());
  mfactors.resize(unique(mfactors.begin(), mfactors.end()) - mfactors.begin());
  vector<long long> phifactors;
  long long phi = etf(m);
  for (long long i = 1; i * i <= phi; i++) {
    if (phi % i == 0) {
      phifactors.push_back(i);
      phifactors.push_back(phi / i);
    }
  }
  sort(phifactors.begin(), phifactors.end());
  phifactors.resize(unique(phifactors.begin(), phifactors.end()) -
                    phifactors.begin());
  long long ans = 1ll;
  for (int i = 1; i < mfactors.size(); i++) {
    long long num = etf(mfactors[i]), den = 0;
    for (auto j : phifactors) {
      if (num % j == 0 && modex(x, j, mfactors[i]) == 1) {
        den = j;
        break;
      }
    }
    ans += num / den;
  }
  printf("%lld\n", ans);
}