#include <bits/stdc++.h>
using namespace std;
using namespace chrono;
void checkpoint1() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
bool comp(pair<long long, long long> &a, pair<long long, long long> &b) {
  if (a.first != b.first)
    return a.first < b.first;
  else {
    if (a.second > b.second)
      return b.second < a.second;
    else if (a.second < b.second)
      return a.second < b.second;
  }
  return a.first > b.first;
}
long long powe(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res *= a;
    }
    a *= a;
    b >>= 1;
  }
  return res;
}
void terminAtor() {
  long long t;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    long long k = p % q;
    long long oo = q;
    if (k != 0)
      cout << p << "\n";
    else {
      map<long long, long long> m;
      for (long long i = 2; i * i <= q; i++) {
        if ((q % i) == 0) {
          while ((q % i) == 0) {
            q /= i;
          }
          m[i] = 1;
        }
      }
      if (q != 1) m[q] = 1;
      long long maxi = 1;
      for (auto i : m) {
        long long ok = i.first, op = p;
        bool ol = 0;
        while ((op % ok) == 0) {
          if ((op % oo) != 0) {
            maxi = max(op, maxi);
            ol = 1;
          }
          op /= ok;
          if (ol) break;
        }
        maxi = max(op, maxi);
      }
      cout << maxi << "\n";
    }
  }
}
int main() {
  checkpoint1();
  auto startrrr = high_resolution_clock::now();
  cout << setprecision(20);
  terminAtor();
  auto stoprrr = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stoprrr - startrrr);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
using namespace chrono;
void checkpoint1() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
bool comp(pair<long long, long long> &a, pair<long long, long long> &b) {
  if (a.first != b.first)
    return a.first < b.first;
  else {
    if (a.second > b.second)
      return b.second < a.second;
    else if (a.second < b.second)
      return a.second < b.second;
  }
  return a.first > b.first;
}
long long powe(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res *= a;
    }
    a *= a;
    b >>= 1;
  }
  return res;
}
void terminAtor() {
  long long t;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    long long k = p % q;
    long long oo = q;
    if (k != 0)
      cout << p << "\n";
    else {
      map<long long, long long> m;
      for (long long i = 2; i * i <= q; i++) {
        if ((q % i) == 0) {
          while ((q % i) == 0) {
            q /= i;
          }
          m[i] = 1;
        }
      }
      if (q != 1) m[q] = 1;
      long long maxi = 1;
      for (auto i : m) {
        long long ok = i.first, op = p;
        bool ol = 0;
        while ((op % ok) == 0) {
          if ((op % oo) != 0) {
            maxi = max(op, maxi);
            ol = 1;
          }
          op /= ok;
          if (ol) break;
        }
        maxi = max(op, maxi);
      }
      cout << maxi << "\n";
    }
  }
}
int main() {
  checkpoint1();
  auto startrrr = high_resolution_clock::now();
  cout << setprecision(20);
  terminAtor();
  auto stoprrr = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stoprrr - startrrr);
  return 0;
}

```