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>
long long Solve(long long p, long long q) {
  if (p % q != 0) return p;
  std::vector<int> primes;
  long long qq = q;
  for (int i = 2; i <= sqrt(qq); i++) {
    if (qq % i == 0) {
      primes.push_back(i);
      while (qq % i == 0) qq /= i;
    }
  }
  if (qq > 1) primes.push_back(qq);
  std::vector<long long> candidates;
  for (const auto prime : primes) {
    long long pp = p;
    while (pp % q == 0) {
      pp /= prime;
    }
    candidates.push_back(pp);
  }
  return *std::max_element(candidates.begin(), candidates.end());
}
int main() {
  int t;
  std::cin >> t;
  while (t--) {
    long long p, q;
    std::cin >> p >> q;
    std::cout << Solve(p, q) << std::endl;
  }
  return 0;
}
