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, ans, pri[1000100], tot, num, a[1001000];
bool vis[10000010];
void pre() {
  for (long long i = 2; i <= 10000000; i++) {
    if (!vis[i]) {
      pri[++tot] = i;
    }
    for (long long j = 1; j <= tot && pri[j] * i <= 10000000; j++) {
      vis[pri[j] * i] = 1;
      if (i % pri[j] == 0) break;
    }
  }
}
long long phi(long long x) {
  long long ans = 1;
  for (long long i = 1; i <= tot && pri[i] * pri[i] <= x; i++) {
    if (x % pri[i] == 0) {
      ans *= (pri[i] - 1);
      x /= pri[i];
      while (x % pri[i] == 0) {
        ans *= pri[i];
        x /= pri[i];
      }
    }
  }
  if (x > 1) ans *= (x - 1);
  return ans;
}
void div(long long x) {
  for (long long i = 1; i <= tot && pri[i] * pri[i] <= x; i++) {
    if (x % pri[i] == 0) {
      a[++num] = pri[i];
      while (x % pri[i] == 0) x /= pri[i];
    }
  }
  if (x > 1) a[++num] = x;
}
long long mul(long long a, long long b, long long M) {
  long long tmp = a * b - (long long)((long double)a * b / M) * M;
  return (tmp % M + M) % M;
}
long long pw(long long x, long long b, long long M) {
  long long res = 1;
  for (; b; b >>= 1, x = mul(x, x, M))
    if (b & 1) res = mul(res, x, M);
  return res;
}
long long getans(long long X, long long M, long long d) {
  long long t = phi(M);
  num = 0;
  div(t);
  long long ans = 1;
  for (register long long i = 1; i <= num; i++) {
    while (t % a[i] == 0 && pw(X, t / a[i], M) == 1) {
      ans *= a[i];
      t /= a[i];
    }
  }
  return ans;
}
signed main() {
  pre();
  scanf("%lld%lld", &M, &X);
  for (register long long i = 1; i <= sqrt(M); i++) {
    if (M % i == 0) {
      ans += getans(X % (M / i), M / i, i);
      if (i * i < M) ans += getans(X % i, i, M / i);
    }
  }
  printf("%lld\n", ans);
  return 0;
}