#include <bits/stdc++.h>
using namespace std;
void factors(vector<long long>& v, long long n) {
  v.emplace_back(n);
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      v.emplace_back(i);
      if (i != n / i) v.emplace_back(n / i);
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cout << setprecision(10) << fixed;
  long long t = 1;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    if (p % q) {
      cout << p << endl;
      ;
    } else {
      vector<long long> v;
      factors(v, q);
      long long ans = 1;
      for (auto i : v) {
        long long temp = p;
        while (temp % i == 0) {
          temp /= i;
          if (temp % q != 0) {
            ans = max(ans, temp);
            break;
          }
        }
      }
      cout << ans << endl;
    }
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
void factors(vector<long long>& v, long long n) {
  v.emplace_back(n);
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      v.emplace_back(i);
      if (i != n / i) v.emplace_back(n / i);
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cout << setprecision(10) << fixed;
  long long t = 1;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    if (p % q) {
      cout << p << endl;
      ;
    } else {
      vector<long long> v;
      factors(v, q);
      long long ans = 1;
      for (auto i : v) {
        long long temp = p;
        while (temp % i == 0) {
          temp /= i;
          if (temp % q != 0) {
            ans = max(ans, temp);
            break;
          }
        }
      }
      cout << ans << endl;
    }
  }
  return 0;
}

```