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;
using ll = long long;
multiset<ll> fac(ll a) {
  multiset<ll> ret;
  for (ll i = 2; i * i <= a; i++) {
    while (a % i == 0) {
      a /= i;
      ret.insert(i);
    }
  }
  if (a > 1) ret.insert(a);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  for (int i = (0); ((1) > 0 ? i < (t) : i > (t)); i += (1)) {
    ll a, m;
    cin >> a >> m;
    multiset<ll> fa = fac(a), fmm = fac(m);
    ll mul = 1;
    for (ll i : fa) {
      auto it = fmm.find(i);
      if (it != fmm.end()) {
        mul *= i;
        fmm.erase(it);
      }
    }
    set<ll> fm;
    for (ll i : fmm) {
      fm.insert(i);
    }
    ll ans = m / mul;
    for (ll i : fm) {
      ans -= ans / i;
    }
    cout << ans << '\n';
  }
}
