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;
template <typename T>
using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
const int INF = 1 << 30;
const long long INFL = 1LL << 60;
const long long mod = 1000000007;
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long eulerPhi(long long n) {
  if (n == 0) return 0;
  long long ans = n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ans -= ans / i;
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) ans -= ans / n;
  return ans;
}
void solve() {
  long long a, b;
  cin >> a >> b;
  long long c = gcd(a, b);
  a /= c;
  b /= c;
  cout << eulerPhi(b) << endl;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  for (int i = 0; i < int(t); i++) {
    solve();
  }
  return 0;
}
