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>
unsigned long long maxprime(unsigned long long a) {
  unsigned long long maxprime = 0;
  while (a % 2 == 0) {
    maxprime = 2;
    a = a / 2;
  }
  for (int i = 3; i <= sqrt(a); i += 2) {
    while (a % i == 0) {
      maxprime = i;
      a = a / i;
    }
  }
  if (a > 2) {
    return a;
  }
  return maxprime;
}
int main() {
  int t;
  std::cin >> t;
  while (t--) {
    unsigned long long p, q;
    std::cin >> p >> q;
    if (p % q == 0) {
      std::unordered_map<unsigned long long, int> factors;
      unsigned long long f = q;
      while (f % 2 == 0) {
        if (factors.find(2) == factors.end()) {
          factors[2] = 1;
        } else {
          factors[2] += 1;
        }
        f = f / 2;
      }
      for (int i = 3; i <= sqrt(q); i += 2) {
        while (f % i == 0) {
          if (factors.find(i) == factors.end()) {
            factors[i] = 1;
          } else {
            factors[i] += 1;
          }
          f = f / i;
        }
      }
      if (f > 2) {
        if (factors.find(f) == factors.end()) {
          factors[f] = 1;
        } else {
          factors[f] += 1;
        }
      }
      unsigned long long max = 1;
      for (auto& r : factors) {
        unsigned long long res = p;
        while (res % r.first == 0) {
          res = res / r.first;
          if (res % q != 0) {
            max = std::max(res, max);
          }
        }
      }
      std::cout << max << std::endl;
    } else {
      std::cout << p << std::endl;
    }
  }
}
