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 = 100;
long long a, b, p[N], q[N];
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> a >> b;
    if (a % b)
      cout << a << endl;
    else {
      p[0] = 0;
      for (int i = 2; i <= sqrt(b); i++) {
        if (b % i == 0) {
          b /= i, p[++p[0]] = i, q[p[0]] = 1;
          while (b % i == 0) q[p[0]]++, b /= i;
        }
      }
      if (b > 1) p[++p[0]] = b, q[p[0]] = 1;
      long long v = a;
      for (int i = 1; i <= p[0]; i++) {
        long long w = a, b = 1, c = 1;
        while (w % p[i] == 0) w /= p[i], b++;
        b -= q[i];
        while (b--) c *= p[i];
        v = min(c, v);
      }
      cout << a / v << endl;
    }
  }
}
