#include <bits/stdc++.h>
using namespace std;
int n;
int main() {
  int t;
  cin >> t;
  long long p;
  int q;
  while (t--) {
    cin >> p >> q;
    if (p % q) {
      cout << p << endl;
      continue;
    }
    vector<int> prime;
    int x = q;
    for (int i = 2; i <= sqrt(q); ++i)
      if (x % i == 0) {
        prime.push_back(i);
        while (x % i == 0) x /= i;
      }
    if (x != 1) prime.push_back(x);
    long long ans = 1;
    for (int fact : prime) {
      long long x = p;
      while (q % fact == 0 && x % q == 0) x /= fact;
      ans = max(ans, x);
    }
    cout << ans << endl;
  }
}

### Prompt

Create a solution in Cpp for the following problem:

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 n;
int main() {
  int t;
  cin >> t;
  long long p;
  int q;
  while (t--) {
    cin >> p >> q;
    if (p % q) {
      cout << p << endl;
      continue;
    }
    vector<int> prime;
    int x = q;
    for (int i = 2; i <= sqrt(q); ++i)
      if (x % i == 0) {
        prime.push_back(i);
        while (x % i == 0) x /= i;
      }
    if (x != 1) prime.push_back(x);
    long long ans = 1;
    for (int fact : prime) {
      long long x = p;
      while (q % fact == 0 && x % q == 0) x /= fact;
      ans = max(ans, x);
    }
    cout << ans << endl;
  }
}

```