You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).

Note: \gcd(a, b) is the greatest common divisor of a and b.

Input

The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.

Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).

Output

Print T integers — one per test case. For each test case print the number of appropriate x-s.

Example

Input


3
4 9
5 10
42 9999999967


Output


6
1
9999999966

Note

In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].

In the second test case the only appropriate x is 0.
#include <bits/stdc++.h>
using namespace std;
vector<long long> p;
long long max(long long a, long long b) { return a > b ? a : b; }
long long min(long long a, long long b) { return a < b ? a : b; }
long long GCD(long long a, long long b) { return b == 0 ? a : GCD(b, a % b); }
long long phi(long long n) {
  long long ans = n;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) {
      ans = ans / i * (i - 1);
      while (n % i == 0) n /= i;
    }
  if (n > 1) ans = ans / n * (n - 1);
  return ans;
}
long long rongchi(long long x) {
  long long ans = 0;
  for (int i = 1; i < (1 << p.size()); i++) {
    int bits = 0;
    long long mul = 1;
    for (int j = 0; j < p.size(); j++)
      if (i & (1 << j)) mul *= p[j], bits++;
    if (bits & 1)
      ans += x / mul;
    else
      ans -= x / mul;
  }
  return ans;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    p.clear();
    long long a, m, gcd;
    scanf("%lld%lld", &a, &m);
    gcd = GCD(a, m);
    if (gcd == 1)
      printf("%lld\n", phi(m));
    else {
      m /= gcd;
      a /= gcd;
      long long tmp = m;
      for (long long i = 2; i * i <= tmp; i++)
        if (tmp % i == 0) {
          p.push_back(i);
          while (tmp % i == 0) tmp /= i;
        }
      if (tmp > 1) p.push_back(tmp);
      printf("%lld\n", m - (rongchi(a + m - 1) - rongchi(a)));
    }
  }
  return 0;
}
