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 PrimeFactorization {
  using number_type = long long;
  using exponent_type = int;

 private:
  map<number_type, exponent_type> factors_;
  number_type n;

 public:
  PrimeFactorization(number_type nn) : n(nn) {
    for (number_type i = 2; i * i <= nn; ++i)
      while (nn % i == 0) ++factors_[i], nn /= i;
    if (nn != 1) ++factors_[nn];
  }
  const map<number_type, exponent_type>& factors() const { return factors_; }
};
template <typename T, typename U>
T& upd_max(T& a, const U& b) {
  return a = max(a, static_cast<T>(b));
}
void solve() {
  using ll = long long;
  ll p, q;
  cin >> p >> q;
  if (p % q)
    cout << p << '\n';
  else {
    PrimeFactorization fq(q);
    ll ans = 1;
    for (const auto& [n, ignore] : fq.factors()) {
      ll t = p;
      while (t % q == 0) t /= n;
      upd_max(ans, t);
    }
    cout << ans << '\n';
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  for (int i = 0; i < t; ++i) solve();
  return 0;
}
