#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long p, q;
  cin >> p >> q;
  if (p % q != 0) {
    cout << p << endl;
    return;
  }
  vector<long long> qs;
  long long q1 = q;
  if (q1 % 2 == 0) {
    qs.push_back(2);
    while (q1 % 2 == 0) q1 /= 2;
  }
  for (long long d = 3; d * d <= q1; d += 2) {
    if (q1 % d == 0) {
      qs.push_back(d);
      while (q1 % d == 0) q1 /= d;
    }
  }
  if (q1 > 1) qs.push_back(q1);
  long long x = 0;
  for (auto& q1 : qs) {
    long long p1 = p;
    while (p1 % q1 == 0) {
      p1 /= q1;
      if (p1 % q != 0) x = max(x, p1);
    }
  }
  cout << x << endl;
}
int main() {
  int tc;
  cin >> tc;
  while (tc--) {
    solve();
  }
}

### 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;
  if (p % q != 0) {
    cout << p << endl;
    return;
  }
  vector<long long> qs;
  long long q1 = q;
  if (q1 % 2 == 0) {
    qs.push_back(2);
    while (q1 % 2 == 0) q1 /= 2;
  }
  for (long long d = 3; d * d <= q1; d += 2) {
    if (q1 % d == 0) {
      qs.push_back(d);
      while (q1 % d == 0) q1 /= d;
    }
  }
  if (q1 > 1) qs.push_back(q1);
  long long x = 0;
  for (auto& q1 : qs) {
    long long p1 = p;
    while (p1 % q1 == 0) {
      p1 /= q1;
      if (p1 % q != 0) x = max(x, p1);
    }
  }
  cout << x << endl;
}
int main() {
  int tc;
  cin >> tc;
  while (tc--) {
    solve();
  }
}

```