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;
const int N = 2e5 + 7;
const int M = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int n, m;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long ans = 0;
long long phi(long long n) {
  long long res = n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) res = res / i * (i - 1);
    while (n % i == 0) n /= i;
  }
  if (n > 1) res = res / n * (n - 1);
  return res;
}
void solve() {
  long long a, m, d;
  scanf("%lld%lld", &a, &m);
  d = gcd(a, m);
  a /= d;
  m /= d;
  ans = phi(m);
  cout << ans << endl;
}
int main() {
  int T;
  scanf("%d", &T);
  for (int i = 1; i <= T; i++) solve();
}
