Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.

To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that: 

  * p_i is divisible by x_i; 
  * x_i is not divisible by q_i. 

Oleg is really good at division and managed to find all the answers quickly, how about you?

Input

The first line contains an integer t (1 ≤ t ≤ 50) — the number of pairs.

Each of the following t lines contains two integers p_i and q_i (1 ≤ p_i ≤ 10^{18}; 2 ≤ q_i ≤ 10^{9}) — the i-th pair of integers.

Output

Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.

One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.

Example

Input


3
10 4
12 6
179 822


Output


10
4
179

Note

For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.

For the second pair, where p_2 = 12 and q_2 = 6, note that 

  * 12 is not a valid x_2, since 12 is divisible by q_2 = 6; 
  * 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6. 

The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
#include <bits/stdc++.h>
using namespace std;
long long GCD(long long A, long long B) { return (B) ? GCD(B, A % B) : A; }
long long LCM(long long A, long long B) { return A * B / GCD(A, B); }
long long max(long long A, long long B) { return (A >= B) ? A : B; }
long long min(long long A, long long B) { return (A <= B) ? A : B; }
bool isPrime(long long n) {
  if (n == 1) return false;
  if (n == 2) return true;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    long long ans = 1;
    for (long long d = 2; d * d <= q; d++)
      if (q % d == 0) {
        long long e = 0;
        while (q % d == 0) {
          e++;
          q /= d;
        }
        long long u = 0;
        long long z = p;
        while (z % d == 0) {
          z /= d;
          u++;
        }
        for (long long i = 0; i < min(e - 1, u); i++) z *= d;
        ans = max(ans, z);
      }
    if (q != 1) {
      long long d = q;
      long long e = 0;
      while (q % d == 0) {
        e++;
        q /= d;
      }
      long long u = 0;
      long long z = p;
      while (z % d == 0) {
        z /= d;
        u++;
      }
      for (long long i = 0; i < min(e - 1, u); i++) z *= d;
      ans = max(ans, z);
    }
    cout << ans << endl;
  }
  return 0;
}
