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 kMaxN = 101;
long long p[kMaxN];
int t, n;
long long x, y, g, ans;
long long Gcd(long long x, long long y) {
  if (!y) {
    return x;
  }
  return Gcd(y, x % y);
}
void Resolve(long long y) {
  n = 0;
  for (long long i = 2; i * i <= y; i++) {
    if (y % i == 0) {
      p[++n] = i;
      while (y % i == 0) {
        y /= i;
      }
    }
  }
  if (y > 1) {
    p[++n] = y;
  }
}
void Search(int i, long long s) {
  if (s >= x + y) {
    return;
  }
  if (i > n) {
    ans += (x + y - 1) / s - (x - 1) / s;
    return;
  }
  Search(i + 1, s);
  Search(i + 1, -s * p[i]);
}
int main() {
  cin >> t;
  while (t--) {
    cin >> x >> y;
    g = Gcd(x, y);
    x /= g;
    y /= g;
    Resolve(y);
    ans = 0;
    Search(1, 1);
    cout << ans << endl;
  }
  return 0;
}
