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;
vector<long long> prime_factors;
long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : x; }
int main(int argc, const char* argv[]) {
  int t;
  cin >> t;
  while (t--) {
    prime_factors.clear();
    long long a, m;
    cin >> a >> m;
    long long d = gcd(m, a);
    a /= d;
    m /= d;
    long long res = m;
    for (long long i = 2; i <= sqrt(m) && m != 1; ++i) {
      long long cnt = 0;
      while (m % i == 0) {
        ++cnt;
        m /= i;
      }
      if (cnt) prime_factors.push_back(i);
    }
    if (m != 1) prime_factors.push_back(m);
    for (int i = 0; i < prime_factors.size(); ++i) {
      res /= prime_factors[i];
      res *= prime_factors[i] - 1;
    }
    cout << res << endl;
  }
  return 0;
}
