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;
long long t, p, q, pre[200000 + 5], k, vis[200000 + 5];
void init() {
  for (long long i = 2; i <= 200000 + 5; i++) {
    if (!vis[i]) {
      vis[i] = 1;
      pre[++k] = i;
      for (long long j = i; j * i <= 200000 + 5; j++) vis[i * j] = 1;
    }
  }
}
int main(void) {
  std::ios::sync_with_stdio(false);
  init();
  cin >> t;
  while (t--) {
    cin >> p >> q;
    if (p % q) {
      cout << p << endl;
      continue;
    }
    long long ans = 1, temp = q;
    for (long long i = 1; i <= k; i++) {
      long long tot = 0, res = 1;
      while (temp % pre[i] == 0) {
        temp /= pre[i];
        tot++;
        res *= pre[i];
      }
      if (!tot) continue;
      res /= pre[i];
      if (p % res == 0) {
        long long temp1 = p;
        while (temp1 % pre[i] == 0) temp1 /= pre[i];
        ans = max(ans, res * temp1);
      }
    }
    if (temp > 1) {
      long long res = 1;
      if (p % res == 0) {
        long long temp1 = p;
        while (temp1 % temp == 0) temp1 /= temp;
        ans = max(ans, res * temp1);
      }
    }
    cout << ans << endl;
  }
  return 0;
}
