#include <bits/stdc++.h>
using namespace std;
vector<pair<int64_t, int64_t> > factorize(long long n) {
  vector<pair<int64_t, int64_t> > v;
  int64_t ans = n, temp;
  int64_t coun = 0;
  while (!(n % 2)) {
    n >>= 1;
    coun++;
  }
  if (coun) {
    v.push_back(make_pair(2, coun));
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    coun = 0;
    while (n % i == 0) {
      coun++;
      n = n / i;
    }
    if (coun) {
      v.push_back(make_pair(i, coun));
    }
  }
  if (n > 2) v.push_back(make_pair(n, 1));
  return v;
}
int32_t main() {
  int64_t i, j, p, q;
  int64_t t;
  cin >> t;
  while (t--) {
    cin >> p >> q;
    if (p % q != 0 || q > p)
      cout << p << '\n';
    else {
      int64_t ans = -1;
      vector<pair<int64_t, int64_t> > temp = factorize(q);
      for (i = 0; i < temp.size(); i++) {
        int64_t tempans = 0;
        int64_t t1 = p;
        while (t1 % temp[i].first == 0) {
          t1 /= temp[i].first;
        }
        int64_t zz = pow(temp[i].first, temp[i].second - 1);
        t1 *= zz;
        ans = max(ans, t1);
      }
      cout << ans << '\n';
    }
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
vector<pair<int64_t, int64_t> > factorize(long long n) {
  vector<pair<int64_t, int64_t> > v;
  int64_t ans = n, temp;
  int64_t coun = 0;
  while (!(n % 2)) {
    n >>= 1;
    coun++;
  }
  if (coun) {
    v.push_back(make_pair(2, coun));
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    coun = 0;
    while (n % i == 0) {
      coun++;
      n = n / i;
    }
    if (coun) {
      v.push_back(make_pair(i, coun));
    }
  }
  if (n > 2) v.push_back(make_pair(n, 1));
  return v;
}
int32_t main() {
  int64_t i, j, p, q;
  int64_t t;
  cin >> t;
  while (t--) {
    cin >> p >> q;
    if (p % q != 0 || q > p)
      cout << p << '\n';
    else {
      int64_t ans = -1;
      vector<pair<int64_t, int64_t> > temp = factorize(q);
      for (i = 0; i < temp.size(); i++) {
        int64_t tempans = 0;
        int64_t t1 = p;
        while (t1 % temp[i].first == 0) {
          t1 /= temp[i].first;
        }
        int64_t zz = pow(temp[i].first, temp[i].second - 1);
        t1 *= zz;
        ans = max(ans, t1);
      }
      cout << ans << '\n';
    }
  }
  return 0;
}

```