#include <bits/stdc++.h>
using namespace std;
void gcd(unsigned long long n,
         unordered_map<unsigned long long, unsigned long long> &hash) {
  while (n % 2 == 0) {
    hash[2]++;
    n = n / 2;
  }
  for (unsigned long long i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      hash[i]++;
      n = n / i;
    }
  }
  if (n > 2) hash[n]++;
  return;
}
int32_t main() {
  unsigned long long t;
  cin >> t;
  for (unsigned long long count = 1; count <= t; count++) {
    unsigned long long p, q;
    cin >> p >> q;
    if (q > p)
      cout << p << endl;
    else {
      if (p % q == 0) {
        unordered_map<unsigned long long, unsigned long long> hash;
        gcd(q, hash);
        unsigned long long mi = 0;
        for (auto it : hash) {
          unsigned long long factor = it.first;
          unsigned long long count = it.second;
          unsigned long long curr = 1;
          unsigned long long temp = p;
          while (temp % q == 0) {
            temp /= factor;
            curr *= factor;
          }
          mi = max(mi, temp);
        }
        cout << mi << endl;
      } else
        cout << p << endl;
    }
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
void gcd(unsigned long long n,
         unordered_map<unsigned long long, unsigned long long> &hash) {
  while (n % 2 == 0) {
    hash[2]++;
    n = n / 2;
  }
  for (unsigned long long i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      hash[i]++;
      n = n / i;
    }
  }
  if (n > 2) hash[n]++;
  return;
}
int32_t main() {
  unsigned long long t;
  cin >> t;
  for (unsigned long long count = 1; count <= t; count++) {
    unsigned long long p, q;
    cin >> p >> q;
    if (q > p)
      cout << p << endl;
    else {
      if (p % q == 0) {
        unordered_map<unsigned long long, unsigned long long> hash;
        gcd(q, hash);
        unsigned long long mi = 0;
        for (auto it : hash) {
          unsigned long long factor = it.first;
          unsigned long long count = it.second;
          unsigned long long curr = 1;
          unsigned long long temp = p;
          while (temp % q == 0) {
            temp /= factor;
            curr *= factor;
          }
          mi = max(mi, temp);
        }
        cout << mi << endl;
      } else
        cout << p << endl;
    }
  }
  return 0;
}

```