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;
const int N = 2e6 + 9;
vector<pair<int, int> > f;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    long long p, q;
    scanf("%lld%lld", &p, &q);
    if (p % q) {
      printf("%lld\n", p);
      continue;
    }
    for (int i = 2; i * i <= q; ++i) {
      if (q % i == 0) {
        int v = i, s = 1;
        while (q % i == 0) {
          s *= i;
          q /= i;
        }
        f.push_back(make_pair(v, s));
      }
    }
    if (q > 1) f.push_back(make_pair(q, q));
    long long ans = 0;
    for (auto w : f) {
      int d = w.second;
      long long P = p;
      while (P % d == 0) {
        P /= w.first;
      }
      ans = max(ans, P);
    }
    printf("%lld\n", ans);
    f.clear();
  }
}
