#include <bits/stdc++.h>
using namespace std;
int main() {
  int64_t t;
  cin >> t;
  while (t--) {
    int64_t p, q;
    cin >> p >> q;
    set<int64_t> s1;
    if (p % q != 0)
      cout << p << endl;
    else {
      int64_t ans, p1 = p, q2 = q;
      while (p1 % q == 0) {
        p1 = p1 / q;
      }
      ans = p1;
      for (int64_t i = 2; i * i <= q2; i++) {
        if (q2 % i != 0) continue;
        if (q2 % i == 0) {
          s1.insert(i);
          while (q2 % i == 0) q2 = q2 / i;
        }
      }
      if (q2 != 1 && q2 != q) s1.insert(q2);
      for (auto x : s1) {
        p1 = p;
        while (p1 % q == 0) {
          p1 = p1 / x;
        }
        ans = max(ans, p1);
      }
      cout << ans << endl;
    }
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int64_t t;
  cin >> t;
  while (t--) {
    int64_t p, q;
    cin >> p >> q;
    set<int64_t> s1;
    if (p % q != 0)
      cout << p << endl;
    else {
      int64_t ans, p1 = p, q2 = q;
      while (p1 % q == 0) {
        p1 = p1 / q;
      }
      ans = p1;
      for (int64_t i = 2; i * i <= q2; i++) {
        if (q2 % i != 0) continue;
        if (q2 % i == 0) {
          s1.insert(i);
          while (q2 % i == 0) q2 = q2 / i;
        }
      }
      if (q2 != 1 && q2 != q) s1.insert(q2);
      for (auto x : s1) {
        p1 = p;
        while (p1 % q == 0) {
          p1 = p1 / x;
        }
        ans = max(ans, p1);
      }
      cout << ans << endl;
    }
  }
}

```