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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long int p, q;
    cin >> p >> q;
    long long int temp = q, div = 2;
    vector<long long int> a, b;
    if (p % q != 0)
      cout << p << "\n";
    else {
      while (temp > 1 && div <= sqrt(temp)) {
        while ((long long int)sqrt(temp) * (long long int)sqrt(temp) == temp) {
          a.push_back(sqrt(temp));
          temp = temp / sqrt(temp);
        }
        if (temp % div == 0) {
          a.push_back(div);
        }
        while (temp % div == 0) {
          temp = temp / div;
        }
        if (div % 2 != 0)
          div += 2;
        else
          div++;
      }
      if (temp > 1) a.push_back(temp);
      if (a.size() == 0)
        cout << "1\n";
      else {
        temp = p;
        for (long long int i = 0; i < a.size(); i++) {
          while (p % a[i] == 0) {
            p /= a[i];
            if (p % q != 0) {
              b.push_back(p);
              break;
            }
          }
          p = temp;
        }
        long long int val = max_element(b.begin(), b.end()) - b.begin();
        cout << b[val] << "\n";
      }
    }
  }
}
