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;
struct node {
  long long int a, b;
};
bool compare(node x, node y) { return x.a > y.a; }
void printDivisors(int n) {
  for (int i = 1; i <= n; i++)
    if (n % i == 0) printf("%d ", i);
}
long long int maxSum(vector<long long int> x) {
  long long int cur_sum = 0;
  long long int max_sum = 0;
  for (long long int i = 0; i < x.size(); i++) {
    cur_sum += x[i];
    if (max_sum < cur_sum) max_sum = cur_sum;
    if (cur_sum < 0) cur_sum = 0;
  }
  return max_sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long long int p, q;
    cin >> p >> q;
    ;
    if (p < q || p % q)
      cout << p << "\n";
    else {
      vector<long long int> v;
      for (long long int i = 1; i <= sqrt(q); i++) {
        if (q % i == 0) {
          v.push_back(i);
          if (i != q / i) v.push_back(q / i);
        }
      }
      long long int res = 1;
      sort(v.begin(), v.end(), greater<long long int>());
      for (long long int i = 0; i < v.size() - 1; i++) {
        long long int x = p;
        while (x % v[i] == 0) {
          if (x < q || x % q)
            break;
          else
            x /= v[i];
        }
        res = max(res, x);
      }
      cout << res << "\n";
    }
  }
  return 0;
}
