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;
inline int ci() {
  register char ch;
  int f = 1;
  while (!isdigit(ch = getchar())) f = ch == '-' ? -1 : 1;
  register int x = ch ^ '0';
  while (isdigit(ch = getchar())) x = (x * 10) + (ch ^ '0');
  return f * x;
}
inline long long gcd(long long a, long long b) {
  return b == 0 ? a : gcd(b, a % b);
}
inline long long phi(long long n) {
  int ed = sqrt(n);
  long long ans = n;
  for (register int i = (2); i <= (ed); i++)
    if (n % i == 0) {
      int cnt = 0;
      while (n % i == 0) cnt++, n /= i;
      ans = ans / i * (i - 1);
    }
  if (n != 1) ans = ans / n * (n - 1);
  return ans;
}
int main() {
  int T = ci();
  while (T--) {
    long long a, n;
    scanf("%lld%lld", &a, &n);
    long long g = gcd(a, n);
    a /= g, n /= g;
    printf("%lld\n", phi(n));
  }
  return 0;
}
