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>
int T;
long long a, m, ans;
long long gcd(long long a, long long b) {
  while (b) a %= b, std::swap(a, b);
  return a;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%lld%lld", &a, &m);
    ans = gcd(a, m), a /= ans, m /= ans, ans = m;
    for (long long i = 2; (long long)i * i <= m; ++i) {
      if (m % i == 0) {
        ans /= i, ans *= (i - 1);
        while (m % i == 0) m /= i;
      }
    }
    if (m > 1) ans /= m, ans *= (m - 1);
    printf("%lld\n", ans);
  }
  return 0;
}
