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;
using ld = long double;
ll gc(ll a, ll b) {
  ll r = a % b;
  while (r) {
    a = b;
    b = r;
    r = a % b;
  }
  return b;
}
int main() {
  ll t;
  cin >> t;
  vector<ll> ch[26];
  while (t--) {
    ll a, m;
    cin >> a >> m;
    ll g = gc(a, m);
    m /= g;
    ll pr = 1;
    for (ll i = 2; i * i <= m; i++) {
      ll pp = 1;
      while (m % i == 0) {
        pp *= i;
        m /= i;
      }
      if (pp > 1) {
        pp /= i;
        pp *= i - 1;
        pr *= pp;
      }
    }
    if (m > 1) {
      pr *= m - 1;
    }
    cout << pr << "\n";
  }
}
