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;
void fact(long long n, map<long long, long long> &res) {
  for (long long i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      n /= i;
      res[i]++;
    }
  }
  if (n != 1) res[n]++;
}
long long bin_pow(long long a, long long b) {
  if (b == 0) return 1;
  long long z = bin_pow(a, b / 2);
  if (b % 2 == 1) return z * z * a;
  return z * z;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  long long q;
  cin >> q;
  while (q--) {
    long long q, p;
    cin >> p >> q;
    if (p % q != 0) {
      cout << p << "\n";
      continue;
    }
    map<long long, long long> mq;
    fact(q, mq);
    long long ans = 0;
    for (auto d : mq) {
      long long k = 0;
      long long x = p;
      while (x % d.first == 0) {
        x /= d.first;
        k++;
      }
      ans =
          max(ans, p / bin_pow(d.first, k) * bin_pow(d.first, (d.second - 1)));
    }
    cout << ans << "\n";
  }
}
