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 a, m, t;
long long fai(long long n) {
  long long sum = n;
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) sum = sum / i * (i - 1);
    while (n % i == 0) n /= i;
  }
  if (n > 1) sum = sum / n * (n - 1);
  return sum;
}
long long gcd(long long x, long long y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
int main() {
  cin >> t;
  while (t--) {
    cin >> a >> m;
    if (gcd(a, m) == 1)
      cout << fai(m) << endl;
    else
      cout << fai(m / gcd(a, m)) << endl;
  }
}
