#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long p, q;
  cin >> p >> q;
  map<long long, int> pm, qm;
  long long p1 = p, q1 = q;
  for (long long i = 2; i * i <= q; i++) {
    if (q % i == 0) {
      while (q % i == 0) {
        q /= i;
        qm[i]++;
      }
    }
  }
  if (q > 1) {
    qm[q]++;
  }
  q = q1;
  long long res = 0;
  for (auto x : qm) {
    int pp = 0;
    while (p % x.first == 0) {
      p /= x.first;
      pp++;
    }
    long long fact = max(pp - x.second + 1, 0);
    p = p1;
    long long curr = 1LL;
    for (int i = 0; i < fact; i++) {
      curr *= x.first;
    }
    p /= curr;
    res = max(res, p);
    p = p1;
  }
  cout << res << "\n";
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

### 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;
void solve() {
  long long p, q;
  cin >> p >> q;
  map<long long, int> pm, qm;
  long long p1 = p, q1 = q;
  for (long long i = 2; i * i <= q; i++) {
    if (q % i == 0) {
      while (q % i == 0) {
        q /= i;
        qm[i]++;
      }
    }
  }
  if (q > 1) {
    qm[q]++;
  }
  q = q1;
  long long res = 0;
  for (auto x : qm) {
    int pp = 0;
    while (p % x.first == 0) {
      p /= x.first;
      pp++;
    }
    long long fact = max(pp - x.second + 1, 0);
    p = p1;
    long long curr = 1LL;
    for (int i = 0; i < fact; i++) {
      curr *= x.first;
    }
    p /= curr;
    res = max(res, p);
    p = p1;
  }
  cout << res << "\n";
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```