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.
#include <bits/stdc++.h>
using namespace std;
long long n, m, ans;
int len, b[100], p[1000005], tot;
bool flag[10000005];
pair<long long, long long> v[100];
long long cf(long long x, long long y, long long M) {
  return (__int128)x * y % M;
}
long long ksm(long long x, long long y, long long M) {
  long long ans = 1;
  for (; y; y >>= 1, x = cf(x, x, M))
    if (y & 1) ans = cf(ans, x, M);
  return ans;
}
void dfs(int x) {
  if (x > len) {
    long long s1 = 1, s2 = 1;
    for (int i = 1; i <= len; i++) {
      for (int j = 1; j <= v[i].second - b[i]; j++) s1 *= v[i].first;
      if (v[i].second != b[i]) {
        s2 *= v[i].first - 1;
        for (int j = 1; j < v[i].second - b[i]; j++) s2 *= v[i].first;
      }
    }
    if (s1 == 1) return;
    long long num = 1, t = s2;
    for (int i = 1; (long long)p[i] * p[i] <= t; i++)
      if (t % p[i] == 0) {
        while (t % p[i] == 0) t /= p[i];
        while (s2 % p[i] == 0 && ksm(m, s2 / p[i], s1) == 1)
          s2 /= p[i], num *= p[i];
      }
    if (t != 1) {
      while (s2 % t == 0 && ksm(m, s2 / t, s1) == 1) s2 /= t, num *= t;
    }
    ans += num;
    return;
  }
  for (int i = 0; i <= v[x].second; i++) {
    b[x] = i;
    dfs(x + 1);
  }
}
int main() {
  scanf("%lld%lld", &n, &m);
  for (int i = 2; i <= 10000000; i++)
    if (!flag[i]) {
      p[++tot] = i;
      for (int j = i; j <= 10000000; j += i) flag[j] = 1;
    }
  long long t = n;
  for (int i = 2; (long long)i * i <= t; i++)
    if (t % i == 0) {
      int cnt = 0;
      while (t % i == 0) t /= i, cnt++;
      v[++len] = make_pair(i, cnt);
    }
  if (t != 1) v[++len] = make_pair(t, 1);
  dfs(1);
  printf("%lld\n", ans + 1);
  return 0;
}
