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 nxt() {
  long long x;
  cin >> x;
  return x;
}
long long _gcd(long long a, long long b) {
  if (a == 0) return b;
  return _gcd(b % a, a);
}
void testcase() {
  long long a, m;
  cin >> a >> m;
  const long long gcd = _gcd(a, m);
  const long long m_dash = m / gcd;
  long long f = (long long)(sqrt(m_dash) + 2);
  vector<long long> prime_factors;
  long long m_dash_2 = m_dash;
  for (long long i = 2; i < f; i++) {
    if (m_dash_2 % i == 0) {
      prime_factors.push_back(i);
      while (m_dash_2 % i == 0) {
        m_dash_2 /= i;
      }
    }
  }
  if (m_dash_2 > 2) {
    prime_factors.push_back(m_dash_2);
  }
  long long ans = m_dash;
  for (long long x : prime_factors) {
    ans -= ans / x;
  }
  cout << ans << "\n";
}
signed main() {
  std::ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) testcase();
}
