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;
using lld = long long int;
using pii = pair<int, int>;
using pll = pair<lld, lld>;
using pli = pair<lld, int>;
using pil = pair<int, lld>;
lld n, m;
lld gcd(lld p, lld q) {
  while (p % q) {
    lld t = p % q;
    p = q;
    q = t;
  }
  return q;
}
lld phi(lld x) {
  lld res = x;
  for (lld i = 2; i <= x / i; i++) {
    if (x % i == 0) {
      while (x % i == 0) x /= i;
      res /= i;
      res *= (i - 1);
    }
  }
  if (x > 1) {
    res /= x;
    res *= (x - 1);
  }
  return res;
}
int main() {
  int i, j, k;
  int t = 1, tv = 0;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld %lld", &n, &m);
    lld g = gcd(n, m);
    m /= g;
    printf("%lld\n", phi(m));
  }
}
