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;
vector<long long> v;
int main() {
  scanf("%d", &T);
  for (int t = 1; t <= T; t++) {
    scanf("%lld %lld", &p, &q);
    if (p % q != 0) {
      printf("%lld\n", p);
      continue;
    }
    for (long long x = 1; x * x <= q; x++) {
      if (q % x == 0) {
        v.push_back(x);
        v.push_back(q / x);
      }
    }
    long long res = 1;
    for (int i = 0; i < v.size(); i++) {
      if (v[i] == 1) {
        continue;
      }
      long long x = p;
      while (x % q == 0 && x % v[i] == 0) {
        x /= v[i];
      }
      if (x % q != 0) {
        res = max(res, x);
      }
    }
    printf("%lld\n", res);
    v.clear();
  }
  return 0;
}
