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;
bool prime[1000001];
vector<long long int> v;
int d[10000000];
void sieve() {
  memset(d, -1, sizeof(d));
  d[0] = 1;
  d[1] = 1;
  prime[0] = true;
  prime[1] = true;
  for (long long i = 2; i <= 1000000; i++) {
    d[i] = i;
    if (prime[i] == false) {
      v.push_back(i);
      for (long long j = i * i; j <= 1000000; j += i) {
        prime[j] = true;
        d[j] = i;
      }
    }
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long powx(long long a, long long p) {
  long long ans = 1;
  for (int i = 0; i < p; i++) ans *= a;
  return ans;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long a, m;
    cin >> a >> m;
    long long g = gcd(a, m);
    m /= g;
    long long ans = m;
    for (long long i = 2; i * i <= m; i++) {
      if (m % i == 0) {
        while (m % i == 0) m /= i;
        ans /= i;
        ans *= (i - 1);
      }
    }
    if (m > 1) {
      ans /= m;
      ans *= (m - 1);
    }
    cout << ans << '\n';
  }
  return 0;
}
