#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve() {
  ll p, q;
  cin >> p >> q;
  ll best = 0;
  set<ll> pf;
  ll cq = q;
  for (ll r = 2; r * r <= cq; r++) {
    while (cq % r == 0) {
      pf.insert(r);
      cq /= r;
    }
  }
  if (cq > 1) pf.insert(cq);
  for (ll r : pf) {
    ll cp = p;
    while (cp % q == 0) cp /= r;
    best = max(best, cp);
  }
  cout << best << '\n';
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) solve();
}

### Prompt

In CPP, your task is to solve the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve() {
  ll p, q;
  cin >> p >> q;
  ll best = 0;
  set<ll> pf;
  ll cq = q;
  for (ll r = 2; r * r <= cq; r++) {
    while (cq % r == 0) {
      pf.insert(r);
      cq /= r;
    }
  }
  if (cq > 1) pf.insert(cq);
  for (ll r : pf) {
    ll cp = p;
    while (cp % q == 0) cp /= r;
    best = max(best, cp);
  }
  cout << best << '\n';
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) solve();
}

```