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 power(long long a, long long b) {
  long long res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long fermat_inv(long long y) { return power(y, 1000000007 - 2ll); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    if (p < q)
      cout << p << "\n";
    else if (gcd(p, q) < q)
      cout << p << "\n";
    else {
      long long ans = 1;
      long long fac = p;
      for (long long i = 2; i * i <= q; i++) {
        int ctr = 0;
        long long val = 1ll;
        while (q % i == 0) {
          q /= i;
          ctr++;
        }
        long long p1 = p;
        int ctr1 = 0;
        while (p1 % i == 0) {
          p1 /= i;
          ctr1++;
          if (ctr1 >= ctr) val *= i;
        }
        if (ctr > 0) fac = min(fac, val);
      }
      if (q > 1) {
        long long p1 = p;
        long long val = 1ll;
        while (p1 % q == 0) {
          p1 /= q;
          val *= q;
        }
        fac = min(fac, val);
      }
      ans = max(ans, p / fac);
      cout << ans << "\n";
    }
  }
  return 0;
}
