#include <bits/stdc++.h>
using namespace std;
const int maxn = 105;
int cnt(long long n, long long p) {
  int res = 0;
  while (n != 0 && n % p == 0) {
    res++;
    n /= p;
  }
  return res;
}
long long pow(long long a, long long b) {
  long long res = 1;
  for (int i = 1; i <= b; i++) res *= a;
  return res;
}
long long max(long long a, long long b) { return a > b ? a : b; }
int main() {
  long long n, t, p, q;
  map<long long, long long> mp;
  cin >> t;
  while (t--) {
    cin >> p >> q;
    if (p % q != 0) {
      printf("%lld\n", p);
      continue;
    } else {
      long long res = 0;
      for (int i = 2; i <= sqrt(q); i++) {
        if (q % i == 0) {
          int cntp = cnt(p, i);
          int cntq = cnt(q, i);
          res = max(res, p / pow(i, cntp - cntq + 1));
          cntp = cnt(p, q / i);
          cntq = cnt(q, q / i);
          res = max(res, p / pow(q / i, cntp - cntq + 1));
        }
      }
      if (res == 0) {
        int cntp = cnt(p, q);
        int cntq = cnt(q, q);
        res = max(res, p / pow(q, cntp - cntq + 1));
      }
      printf("%lld\n", res);
    }
  }
  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;
const int maxn = 105;
int cnt(long long n, long long p) {
  int res = 0;
  while (n != 0 && n % p == 0) {
    res++;
    n /= p;
  }
  return res;
}
long long pow(long long a, long long b) {
  long long res = 1;
  for (int i = 1; i <= b; i++) res *= a;
  return res;
}
long long max(long long a, long long b) { return a > b ? a : b; }
int main() {
  long long n, t, p, q;
  map<long long, long long> mp;
  cin >> t;
  while (t--) {
    cin >> p >> q;
    if (p % q != 0) {
      printf("%lld\n", p);
      continue;
    } else {
      long long res = 0;
      for (int i = 2; i <= sqrt(q); i++) {
        if (q % i == 0) {
          int cntp = cnt(p, i);
          int cntq = cnt(q, i);
          res = max(res, p / pow(i, cntp - cntq + 1));
          cntp = cnt(p, q / i);
          cntq = cnt(q, q / i);
          res = max(res, p / pow(q / i, cntp - cntq + 1));
        }
      }
      if (res == 0) {
        int cntp = cnt(p, q);
        int cntq = cnt(q, q);
        res = max(res, p / pow(q, cntp - cntq + 1));
      }
      printf("%lld\n", res);
    }
  }
  return 0;
}

```