#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int p, q;
    cin >> p >> q;
    if (p % q) {
      cout << p << "\n";
    } else {
      vector<pair<long long int, long long int> > candidates;
      for (int i = 2; i * i <= q; i++) {
        if (q % i == 0) {
          long long int count = 0;
          while (q % i == 0) {
            q /= i;
            count++;
          }
          candidates.push_back(make_pair(i, count));
        }
      }
      if (q > 1) {
        candidates.push_back(make_pair(q, 1));
      }
      long long int result = 1;
      for (int i = 0; i < candidates.size(); i++) {
        long long int x = candidates[i].first, count = candidates[i].second,
                      fake_p = p;
        while (fake_p % x == 0) {
          fake_p /= x;
        }
        for (int i = 1; i < count; i++) {
          fake_p *= x;
        }
        result = max(result, fake_p);
      }
      cout << result << "\n";
    }
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to 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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int p, q;
    cin >> p >> q;
    if (p % q) {
      cout << p << "\n";
    } else {
      vector<pair<long long int, long long int> > candidates;
      for (int i = 2; i * i <= q; i++) {
        if (q % i == 0) {
          long long int count = 0;
          while (q % i == 0) {
            q /= i;
            count++;
          }
          candidates.push_back(make_pair(i, count));
        }
      }
      if (q > 1) {
        candidates.push_back(make_pair(q, 1));
      }
      long long int result = 1;
      for (int i = 0; i < candidates.size(); i++) {
        long long int x = candidates[i].first, count = candidates[i].second,
                      fake_p = p;
        while (fake_p % x == 0) {
          fake_p /= x;
        }
        for (int i = 1; i < count; i++) {
          fake_p *= x;
        }
        result = max(result, fake_p);
      }
      cout << result << "\n";
    }
  }
  return 0;
}

```