#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
long long T, P, Q;
void solve() {
  long long ans = 0;
  cin >> P >> Q;
  long long p, q = Q;
  for (long long i = 2; i * i <= Q; i++) {
    if (q % i) continue;
    p = P;
    while (q % i == 0) q /= i;
    while (p % Q == 0) p /= i;
    ans = max(ans, p);
  }
  if (q != 1) {
    p = P;
    while (p % q == 0) p /= q;
    ans = max(ans, p);
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  while (T--) solve();
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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;
const int INF = 1e9;
long long T, P, Q;
void solve() {
  long long ans = 0;
  cin >> P >> Q;
  long long p, q = Q;
  for (long long i = 2; i * i <= Q; i++) {
    if (q % i) continue;
    p = P;
    while (q % i == 0) q /= i;
    while (p % Q == 0) p /= i;
    ans = max(ans, p);
  }
  if (q != 1) {
    p = P;
    while (p % q == 0) p /= q;
    ans = max(ans, p);
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  while (T--) solve();
  return 0;
}

```