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 n;
int m;
int p[100], a[100];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%I64d %d", &n, &m);
    if (n % m) {
      printf("%I64d\n", n);
      continue;
    }
    int tot = 0;
    for (int x = 2; 1ll * x * x <= m; x++) {
      if (m % x) continue;
      p[tot] = x, a[tot] = 0;
      while (m % x == 0) m /= x, a[tot]++;
      tot++;
    }
    if (m > 1) p[tot] = m, a[tot] = 1, tot++;
    long long ans = n, d = n;
    for (int i = 0; i < tot; i++) {
      int cnt = 0;
      while (n % p[i] == 0) n /= p[i], cnt++;
      assert(cnt >= a[i]);
      long long pw = 1;
      while (cnt >= a[i]) cnt--, pw *= p[i];
      d = min(d, pw);
    }
    printf("%I64d\n", ans / d);
  }
  return 0;
}
