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;
vector<pair<long long, int> > v;
vector<pair<long long, int> > u;
map<long long, long long> phi;
vector<long long> dfdfd;
long long product(long long p, long long q, long long mod) {
  if (q == 0) return 0;
  if (q & 1) return (product(p, q - 1, mod) + p) % mod;
  return (product(p, q >> 1, mod) << 1) % mod;
}
long long modpow(long long x, long long y, long long mod) {
  if (y == 0) {
    return 1;
  }
  if (y & 1) return product(modpow(x, y - 1, mod), x, mod);
  long long m = modpow(x, y >> 1, mod);
  return product(m, m, mod);
}
void solve(int i, long long val, long long totient) {
  if (i == v.size()) {
    phi[val] = totient;
    dfdfd.emplace_back(val);
    return;
  }
  long long cur = 1;
  for (int x = 0; x <= v[i].second; x++) {
    long long mult = cur == 1 ? 1 : (cur / v[i].first) * (v[i].first - 1);
    solve(i + 1, val * cur, totient * mult);
    cur *= v[i].first;
  }
}
long long ans;
vector<long long> can;
vector<long long> modulo;
void solver(int x, long long val, int ptr) {
  if (x == v.size()) {
    if (val == 1) return;
    ans += phi[val] / can[ptr];
    return;
  }
  for (int i = 0; i <= v[x].second; i++) {
    while (val > 1 && modulo[ptr] % val != 1) {
      ++ptr;
    }
    solver(x + 1, val, ptr);
    val *= v[x].first;
  }
}
int main(int argc, char const *argv[]) {
  cin >> m >> x;
  long long tmp = m;
  for (long long i = 2; i * i <= m; i++) {
    int cnt = 0;
    while (tmp % i == 0) {
      tmp /= i;
      ++cnt;
    }
    if (cnt) {
      v.emplace_back(i, cnt);
    }
  }
  if (tmp > 1) v.emplace_back(tmp, 1);
  solve(0, 1, 1);
  tmp = phi[m];
  for (long long i = 1; i * i <= tmp; i++) {
    if (tmp % i == 0) {
      can.emplace_back(i);
      if ((i * i) != tmp) {
        can.emplace_back(tmp / i);
      }
    }
  }
  sort(can.begin(), can.end());
  for (auto i : can) {
    modulo.emplace_back(modpow(x, i, m));
  }
  ans = 1;
  solver(0, 1, 0);
  printf("%lld\n", ans);
  return 0;
}