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>
const int maxn = 1e5 + 10;
using namespace std;
long long quickqow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans *= a;
    a *= a;
    b >>= 1;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long euler_phi(long long n) {
  long long m = int(sqrt(n + 0.5));
  long long ans = n;
  for (int i = 2; i <= m; 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;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    long long a, m;
    cin >> a >> m;
    long long p = gcd(a, m);
    m /= p;
    cout << euler_phi(m) << endl;
  }
  return 0;
}
