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 gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
vector<long long> fen(long long m) {
  vector<long long> ret;
  for (int i = 2; (long long)i * i <= m; i++) {
    if (m % i == 0) {
      ret.push_back(i);
      while (m % i == 0) {
        m /= i;
      }
    }
  }
  if (m != 1) {
    ret.push_back(m);
  }
  return ret;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long a, m;
    cin >> a >> m;
    long long d = gcd(a, m);
    m /= d;
    vector<long long> divsor = fen(m);
    vector<pair<long long, int> > p;
    p.push_back(make_pair(1, 0));
    for (int i = 0; i < divsor.size(); i++) {
      long long b = divsor[i];
      int n = p.size();
      for (int j = 0; j < n; j++) {
        p.push_back(make_pair(p[j].first * b, p[j].second ^ 1));
      }
    }
    long long ret = 0;
    for (int i = 0; i < p.size(); i++) {
      if (p[i].second == 0) {
        ret += m / p[i].first;
      } else {
        ret -= m / p[i].first;
      }
    }
    cout << ret << endl;
  }
  return 0;
}
