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 T, x, y, ans, now, s, p[3000010], bz[3000010];
void R(long long &x) {
  x = 0;
  int w = 1;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= w;
}
int main() {
  for (register long long i = 2; i <= 3000000; i++) {
    if (!bz[i]) p[++p[0]] = i;
    for (register long long j = 1; j <= p[0]; j++) {
      if (i * p[j] > 3000000) break;
      bz[i * p[j]] = 1;
      if (i % p[j] == 0) break;
    }
  }
  R(T);
  while (T--) {
    R(x), R(y);
    s = y;
    if (x % y)
      printf("%lld\n", x);
    else {
      ans = 0;
      for (register long long i = 1; i <= p[0]; i++) {
        if (p[i] > y) break;
        if (y % p[i] == 0) {
          while (y % p[i] == 0) y /= p[i];
          now = x;
          while (now % s == 0) now /= p[i];
          ans = max(ans, now);
        }
      }
      if (y > 1) {
        now = x;
        while (now % s == 0) now /= y;
        ans = max(ans, now);
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}
