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 MAXN = 10000;
const long long int MOD = 1000000007;
const int INF = 1000000000;
long long int phi(long long int n) {
  long long int res = n;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      res /= i;
      res *= (i - 1);
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) {
    res /= n;
    res *= (n - 1);
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0ll)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long int a, b, t;
  cin >> t;
  while (t--) {
    cin >> a >> b;
    cout << phi(b / gcd(a, b)) << endl;
  }
  return 0;
}
