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;
int T;
long long a, b, g, maxi;
vector<long long> query;
vector<long long> primes;
vector<long long> factors;
long long gcd(long long x, long long y) {
  if (x == 0) return y;
  if (x > y) return gcd(y, x);
  return gcd(y % x, x);
}
int main() {
  cin >> T;
  for (int i = 0; i < T; i++) {
    cin >> a >> b;
    g = gcd(a, b);
    b /= g;
    query.push_back(b);
  }
  maxi = query[0];
  for (int i = 1; i < T; i++)
    if (maxi < query[i]) maxi = query[i];
  for (long long i = 2; i <= pow(maxi, 0.5) + 2; i++) {
    bool isPrime = 1;
    for (int j = 0; j < primes.size(); j++) {
      if (i % primes[j] == 0) {
        isPrime = 0;
        break;
      }
    }
    if (isPrime) primes.push_back(i);
  }
  for (int i = 0; i < query.size(); i++) {
    b = query[i];
    factors.clear();
    for (int j = 0; j < primes.size(); j++) {
      a = primes[j];
      if (b % a == 0) factors.push_back(a);
      while (b % a == 0) b /= a;
    }
    if (b > primes[primes.size() - 1]) factors.push_back(b);
    b = query[i];
    for (int j = 0; j < factors.size(); j++)
      b = b / factors[j] * (factors[j] - 1);
    cout << b << '\n';
  }
  return 0;
}
