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;
const int mod = 1e9 + 7, A = 2e5 + 5;
long long gcd(long long x, long long y) { return (!y) ? x : gcd(y, x % y); }
long long phi_function(long long n) {
  long long ans = 1;
  for (long long i = 2; i * i <= n; i++) {
    int p = 1;
    while (n % i == 0) n /= i, p *= i;
    ans *= (p - p / i);
  }
  if (n > 1) ans *= (n - 1);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long a, m;
    cin >> a >> m;
    cout << phi_function(m / gcd(a, m)) << '\n';
  }
  return 0;
}
