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 T;
long long p, q;
long long max(long long A, long long B) { return A > B ? A : B; }
long long min(long long A, long long B) { return A < B ? A : B; }
int main() {
  scanf("%d", &T);
  while (T--) {
    long long ans = -1;
    scanf("%lld %lld", &p, &q);
    if (p % q) {
      printf("%lld\n", p);
      continue;
    }
    for (int i = 1; i * i <= q; i++) {
      if (q % i) continue;
      long long tmp = p, cnt = 1;
      if (i > 1) {
        for (; tmp % i == 0; tmp /= i) cnt++;
        for (tmp = q; tmp % i == 0; tmp /= i) cnt--;
        for (tmp = 1; cnt; cnt--) tmp *= i;
        ans = max(ans, p / tmp);
      }
      tmp = p;
      cnt = 1;
      for (; tmp % (q / i) == 0; tmp /= (q / i)) cnt++;
      for (tmp = q; tmp % (q / i) == 0; tmp /= (q / i)) cnt--;
      for (tmp = 1; cnt; cnt--) tmp *= (q / i);
      ans = max(ans, p / tmp);
    }
    printf("%lld\n", ans);
  }
  return 0;
}
