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) { return b ? gcd(b, a % b) : a; }
void solve();
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << setprecision(10);
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
long long a, b;
long long g;
void solve() {
  ;
  cin >> a >> b;
  g = gcd(a, b);
  long long k2 = b / g;
  set<long long> primefacs;
  long long t = k2;
  (k2);
  for (long long i = 2; i * i <= k2; i++) {
    while (t % i == 0) {
      primefacs.insert(i);
      t = t / i;
    }
  }
  if (t != 1) primefacs.insert(t);
  (primefacs);
  long long tot = k2;
  for (auto e : primefacs) {
    tot = (tot / e) * (e - 1);
  }
  {
    cout << tot << "\n";
    return;
  };
}
