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;
template <class T>
void aIn(T *a, int n) {
  for (int i = 0; i < n; i++) std::cin >> a[i];
}
template <class T>
void aOut(T *a, int n) {
  for (int i = 0; i < n; i++) std::cout << a[i] << " ";
  std::cout << endl;
}
template <class T>
void vOut(vector<T> &a) {
  for (T &x : a) std::cout << x << " ";
  std::cout << endl;
}
long long p, q;
long long gen(long long div) {
  long long mul = 1;
  long long tq = q;
  while (tq % div == 0) {
    tq /= div;
    mul *= div;
  }
  mul /= div;
  long long r = p;
  while (r % div == 0) r /= div;
  r = r * mul;
  return r;
}
void solv() {
  cin >> p >> q;
  if (p % q) {
    cout << p << endl;
    return;
  }
  long long ans = -1ll;
  bool prime = 1;
  for (long long x = 2; x * x <= q; x++) {
    if (q % x) continue;
    prime = 0;
    ans = max(ans, gen(x));
    ans = max(ans, gen(q / x));
  }
  if (prime) {
    ans = gen(q);
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    solv();
  }
}
