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;
vector<long long> prime_factors(long long N) {
  int l = 0;
  while (l * l <= N) l++;
  l++;
  vector<char> is_prime(l, 1);
  vector<long long> res;
  res.reserve(l);
  is_prime[0] = is_prime[1] = 0;
  int f = 2;
  while (f * f <= N) {
    if (is_prime[f]) {
      if (N % f == 0) {
        res.push_back(f);
        while (N % f == 0) N /= f;
      }
      for (int j = f * f; j < l; j += f) {
        is_prime[j] = 0;
      }
    }
    f++;
  }
  if (N > 1) res.push_back(N);
  return res;
}
void solve() {
  using ll = long long;
  ll P, Q;
  cin >> P >> Q;
  vector<ll> factors = prime_factors(Q);
  ll ans = 0;
  for (ll i : factors) {
    ll x = P;
    while (x % Q == 0) x /= i;
    ans = max(ans, x);
  }
  cout << ans << '\n';
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
}
