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;
inline long long read() {
  long long x = 0, y = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') y = -y;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * y;
}
namespace qiqi {
void main() {
  long long a = read(), b = read();
  if (a % b) {
    printf("%lld\n", a);
    return;
  }
  long long x = 0, y;
  for (int i = (2); i <= (sqrt(b)); ++i)
    if (!(b % i)) {
      y = a;
      while (!(y % b)) y /= i;
      x = max(x, y);
      y = a;
      while (!(y % b)) y /= b / i;
      x = max(x, y);
    }
  y = a;
  while (!(y % b)) y /= b;
  x = max(x, y);
  printf("%lld\n", x);
}
}  // namespace qiqi
int main() {
  int T = read();
  while (T--) qiqi::main();
  return 0;
}
