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;
const int inf = 1e9 + 5;
const long long int inf64 = 1e18 + 5;
vector<long long int> get_fact(long long int x) {
  vector<long long int> fact;
  for (int i = 2; i <= sqrt(x); i++) {
    if (x % i == 0) fact.push_back(i);
    while (x % i == 0) x /= i;
  }
  if (x > 1) fact.push_back(x);
  return fact;
}
long long int calc(long long int x, long long int p) {
  long long int res = 1;
  while (p--) res *= x;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long int t, a, b;
  cin >> t;
  while (t--) {
    cin >> a >> b;
    if (a < b)
      cout << a << '\n';
    else {
      if (a % b)
        cout << a << '\n';
      else {
        long long int ans = 1;
        vector<long long int> f1 = get_fact(b);
        for (long long int x : f1) {
          long long int p1 = 0, p2 = 0;
          long long int na = a, nb = b;
          while (na % x == 0) {
            ++p1;
            na /= x;
          }
          while (nb % x == 0) {
            ++p2;
            nb /= x;
          }
          long long int rem = p1 - p2 + 1;
          ans = max(ans, a / calc(x, rem));
        }
        cout << ans << '\n';
      }
    }
  }
}
