#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(0);
  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 << "\n";
      continue;
    }
    vector<pair<long long, long long>> f;
    for (long long i = 2; i * i <= q; i++) {
      if (q % i) continue;
      long long cnt = 0;
      while (q % i == 0) {
        cnt++;
        q /= i;
      }
      f.push_back({i, cnt});
    }
    if (q > 1) f.push_back({q, 1});
    long long mx = 1;
    for (pair<long long, long long> x : f) {
      long long cp = p;
      while (cp % x.first == 0) cp /= x.first;
      for (long long i = 0; i < x.second - 1; i++) cp *= x.first;
      mx = max(mx, cp);
    }
    cout << mx << "\n";
  }
  return 0;
}

### Prompt

Please formulate 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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(0);
  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 << "\n";
      continue;
    }
    vector<pair<long long, long long>> f;
    for (long long i = 2; i * i <= q; i++) {
      if (q % i) continue;
      long long cnt = 0;
      while (q % i == 0) {
        cnt++;
        q /= i;
      }
      f.push_back({i, cnt});
    }
    if (q > 1) f.push_back({q, 1});
    long long mx = 1;
    for (pair<long long, long long> x : f) {
      long long cp = p;
      while (cp % x.first == 0) cp /= x.first;
      for (long long i = 0; i < x.second - 1; i++) cp *= x.first;
      mx = max(mx, cp);
    }
    cout << mx << "\n";
  }
  return 0;
}

```