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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long aa, bb;
    cin >> aa >> bb;
    if (aa % bb != 0) {
      cout << aa << endl;
      continue;
    }
    long long ans = 1;
    vector<long long> jj;
    for (long long i = 1; i * i <= bb; i++) {
      if (bb % i == 0) {
        jj.push_back(bb / i);
        jj.push_back(i);
      }
    }
    for (auto i : jj) {
      if (i == 1) {
        continue;
      }
      long long st = 0;
      for (long long j = 2; j * j <= i; j++) {
        if (i % j == 0) {
          st = 1;
          break;
        }
      }
      if (st) {
        continue;
      }
      long long co = 0;
      long long dd = aa;
      while (dd % i == 0) {
        co++;
        dd /= i;
      }
      long long co2 = 0;
      dd = bb;
      while (dd % i == 0) {
        co2++;
        dd /= i;
      }
      long long kk = 1;
      for (long long j = 0; j < max(co - co2 + 1, (long long)0); j++) {
        kk *= i;
      }
      ans = max(ans, aa / (kk));
    }
    cout << ans << endl;
  }
  return 0;
}
