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;
long long gcd(long long a, long long b) {
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
long long phi(long long md) {
  long long result = md;
  for (long long p = 2; p * p <= md; p++) {
    if (md % p == 0) {
      while (md % p == 0) {
        md = md / p;
      }
      result = result / p * (p - 1);
    }
  }
  if (md > 1) {
    result = result / md * (md - 1);
  }
  return result;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long a, m;
    scanf("%lld%lld", &a, &m);
    printf("%lld\n", phi(m / (gcd(a, m))));
  }
}
