#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  if (!b) return 1;
  long long aa = power(a, b / 2);
  if (b % 2) return aa * aa * a;
  return aa * aa;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    if (p % q != 0) {
      cout << p << endl;
      continue;
    }
    map<long long, long long> m;
    long long tem = q;
    while (tem % 2 == 0) {
      m[2]++;
      tem /= 2;
    }
    for (long long i = 3; i <= sqrt(tem); i += 2) {
      while (tem % i == 0) {
        m[i]++;
        tem /= i;
      }
      if (tem == 1) break;
    }
    if (tem > 2) m[tem]++;
    long long ans = 0;
    for (auto a : m) {
      long long c = 0;
      long long temp = p;
      while (temp % a.first == 0) {
        c++;
        temp /= a.first;
      }
      ans = max(ans, p / power(a.first, c - a.second + 1));
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
long long power(long long a, long long b) {
  if (!b) return 1;
  long long aa = power(a, b / 2);
  if (b % 2) return aa * aa * a;
  return aa * aa;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    if (p % q != 0) {
      cout << p << endl;
      continue;
    }
    map<long long, long long> m;
    long long tem = q;
    while (tem % 2 == 0) {
      m[2]++;
      tem /= 2;
    }
    for (long long i = 3; i <= sqrt(tem); i += 2) {
      while (tem % i == 0) {
        m[i]++;
        tem /= i;
      }
      if (tem == 1) break;
    }
    if (tem > 2) m[tem]++;
    long long ans = 0;
    for (auto a : m) {
      long long c = 0;
      long long temp = p;
      while (temp % a.first == 0) {
        c++;
        temp /= a.first;
      }
      ans = max(ans, p / power(a.first, c - a.second + 1));
    }
    cout << ans << endl;
  }
  return 0;
}

```