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 long long N = 20 + 5;
long long T, p, q, x, cnt, ans, a[N], b[N];
long long read() {
  char c;
  long long x = 0, f = 1;
  c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
signed main() {
  T = read();
  while (T--) {
    cnt = 0, memset(a, 0, sizeof(a)), memset(b, 0, sizeof(b));
    p = read(), q = read();
    for (long long i = 2; i * i <= q; ++i)
      if (q % i == 0) {
        a[++cnt] = i;
        while (q % i == 0) ++b[cnt], q /= i;
      }
    if (q != 1) a[++cnt] = q, b[cnt] = 1;
    ans = 0;
    for (long long i = 1; i <= cnt; ++i) {
      long long tmp = 1;
      for (long long j = 1; j <= b[i]; ++j) tmp = tmp * a[i];
      if (p % tmp == 0) {
        long long x = p;
        while (x % a[i] == 0) x /= a[i];
        ans = max(ans, x * tmp / a[i]);
      } else
        ans = max(ans, p);
    }
    printf("%lld\n", ans);
  }
  return 0;
}
