#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 << endl;
    else {
      long long int k = sqrt(q), i = 2, x = q;
      vector<pair<long long int, int> > prime;
      while (i <= k) {
        int count = 0;
        while (x % i == 0) {
          x = x / i;
          count++;
        }
        if (count) prime.push_back(make_pair(i, count));
        i++;
      }
      if (x > 1) prime.push_back(make_pair(x, 1));
      vector<int> count(prime.size());
      x = p;
      for (int i = 0; i < prime.size(); i++) {
        while (x % prime[i].first == 0) {
          count[i]++;
          x = x / prime[i].first;
        }
      }
      int n = prime.size();
      long long int res = 0;
      for (int i = 0; i < prime.size(); i++) {
        int k = count[i] - prime[i].second + 1;
        x = pow(prime[i].first, k);
        res = max(res, p / x);
      }
      cout << res << endl;
    }
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 << endl;
    else {
      long long int k = sqrt(q), i = 2, x = q;
      vector<pair<long long int, int> > prime;
      while (i <= k) {
        int count = 0;
        while (x % i == 0) {
          x = x / i;
          count++;
        }
        if (count) prime.push_back(make_pair(i, count));
        i++;
      }
      if (x > 1) prime.push_back(make_pair(x, 1));
      vector<int> count(prime.size());
      x = p;
      for (int i = 0; i < prime.size(); i++) {
        while (x % prime[i].first == 0) {
          count[i]++;
          x = x / prime[i].first;
        }
      }
      int n = prime.size();
      long long int res = 0;
      for (int i = 0; i < prime.size(); i++) {
        int k = count[i] - prime[i].second + 1;
        x = pow(prime[i].first, k);
        res = max(res, p / x);
      }
      cout << res << endl;
    }
  }
  return 0;
}

```