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;
long long read() {
  long long s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x < 10)
    putchar(x + '0');
  else {
    write(x / 10);
    putchar(x % 10 + '0');
  }
}
void print(long long x, char c = '\n') {
  write(x);
  putchar(c);
}
signed main() {
  long long T = read();
  while (T--) {
    long long x = read(), y = read(), ans = 0;
    if (x % y > 0)
      print(x);
    else {
      long long s = sqrt(y), tmpy = y, tmpx;
      for (long long i = 2; i <= s; i++)
        if (tmpy % i == 0) {
          while (tmpy % i == 0) tmpy /= i;
          tmpx = x;
          while (tmpx % y == 0) tmpx /= i;
          ans = max(ans, tmpx);
        }
      if (tmpy > 1) {
        tmpx = x;
        while (tmpx % y == 0) tmpx /= tmpy;
        ans = max(ans, tmpx);
      }
      print(ans);
    }
  }
  return 0;
}
