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 long long B = (long long)1e4;
long long mult(long long x, long long y, long long mod) {
  if (y == 0) return 0;
  return (mult(x, y / B, mod) * B + x * (y % B)) % mod;
}
long long bin_pow(long long x, long long p, long long mod) {
  if (p == 0) return 1 % mod;
  if (p & 1) return mult(x, bin_pow(x, p - 1, mod), mod);
  return bin_pow(mult(x, x, mod), p / 2, mod);
}
long long gcd(long long x, long long y) { return y == 0 ? x : gcd(y, x % y); }
long long lcm(long long x, long long y) { return x * (y / gcd(x, y)); }
const int N = 50;
int n;
long long P[N][2];
long long a[N][N][2];
vector<pair<long long, long long> > factorize(long long m) {
  vector<pair<long long, long long> > res;
  for (long long y = 2; y * y <= m; y++) {
    if (m % y) continue;
    res.push_back(make_pair(y, 0LL));
    while (m % y == 0) {
      res.back().second++;
      m /= y;
    }
    n++;
  }
  if (m > 1) {
    res.push_back(make_pair(m, 1LL));
  }
  return res;
}
long long ans;
void brute(int pos, long long PP, long long Q) {
  if (pos == n) {
    ans += PP / Q;
    return;
  }
  for (int i = 0; i <= P[pos][1]; i++) {
    brute(pos + 1, PP * a[pos][i][0], lcm(Q, a[pos][i][1]));
  }
}
int main() {
  long long m, x;
  cin >> m >> x;
  vector<pair<long long, long long> > q = factorize(m);
  n = (int)q.size();
  for (int i = 0; i < n; i++) {
    P[i][0] = q[i].first;
    P[i][1] = q[i].second;
  }
  ans = 0;
  for (int i = 0; i < n; i++) {
    long long mod = 1;
    q = factorize(P[i][0] - 1);
    for (int pw = 0; pw <= P[i][1]; pw++) {
      if (pw == 0) {
        a[i][pw][0] = a[i][pw][1] = 1;
        continue;
      }
      mod *= P[i][0];
      long long cur = mod - mod / P[i][0];
      for (int j = 0; j < (int)q.size(); j++) {
        for (int k = 0; k < q[j].second; k++) {
          long long ncur = cur / q[j].first;
          if (bin_pow(x, ncur, mod) != 1) break;
          cur = ncur;
        }
      }
      for (int k = 0; k < pw - 1; k++) {
        long long ncur = cur / P[i][0];
        if (bin_pow(x, ncur, mod) != 1) break;
        cur = ncur;
      }
      a[i][pw][0] = mod - mod / P[i][0];
      a[i][pw][1] = cur;
    }
  }
  brute(0, 1, 1);
  printf("%lld\n", ans);
  return 0;
}