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;
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
vector<long long> p;
long long is_prime[100010];
void crivo() {
  for (int i = 2; i < 100010; i++) {
    if (!is_prime[i]) {
      p.push_back(i);
      for (int j = 2 * i; j < 100010; j += i) {
        is_prime[j] = 1;
      }
    }
  }
}
long long phi(long long x) {
  long long aux = x;
  int it = 0;
  while (aux > 1 && it < p.size()) {
    if (aux % p[it] == 0) {
      x /= p[it];
      x *= p[it] - 1;
      while (aux % p[it] == 0) aux /= p[it];
    }
    it++;
  }
  if (aux > 1) {
    x /= aux;
    x *= aux - 1;
  }
  return x;
}
int main() {
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  crivo();
  while (t--) {
    long long a, m;
    cin >> a >> m;
    long long d = gcd(a, m);
    a /= d;
    m /= d;
    cout << phi(m) << endl;
  }
}
