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.
#include <bits/stdc++.h>
using namespace std;
void solve_test_case() {
  long long p;
  int q;
  cin >> p >> q;
  vector<pair<int, int>> qfactors;
  for (int i = 2; i * i <= q; i++) {
    int count = 0;
    while (q % i == 0) {
      q /= i;
      count++;
    }
    if (count) {
      qfactors.emplace_back(i, count);
    }
  }
  if (q > 1) {
    qfactors.emplace_back(q, 1);
  }
  long long divide_by_min = p;
  for (auto& fcnt : qfactors) {
    int factor = fcnt.first;
    int qcount = fcnt.second;
    int pcount = 0;
    long long pp = p;
    while (pp % factor == 0) {
      pp /= factor;
      pcount++;
    }
    long long divide_by = 1;
    for (int i = 0; i < pcount - qcount + 1; i++) {
      divide_by *= factor;
    }
    divide_by_min = min(divide_by_min, divide_by);
  }
  cout << (p / divide_by_min) << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  int tests;
  cin >> tests;
  for (int t = 0; t < tests; t++) {
    solve_test_case();
  }
  return 0;
}
