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;
vector<long long> sv;
long long ar[1000000 + 9] = {0};
void seiv() {
  long long n = 1000000, i, j;
  ar[1] = 1;
  for (i = 4; i <= n; i += 2) ar[i] = 1;
  for (i = 3; i <= n; i += 2) {
    if (ar[i] == 0) {
      for (j = i * i; j <= n; j += i * 2) ar[j] = 1;
    }
  }
  for (i = 2; i <= n; i++) {
    if (ar[i] == 0) sv.push_back(i);
  }
}
int main() {
  long long t;
  seiv();
  cin >> t;
  while (t--) {
    long long a, b, c, d, e, i, j, k, l = 1;
    cin >> a >> b;
    if (a % b != 0)
      cout << a << "\n";
    else {
      vector<long long> v;
      for (i = 0; i < sv.size(); i++) {
        if (b % sv[i] == 0) v.push_back(sv[i]);
        if (sv[i] > b) break;
      }
      l = 1;
      v.push_back(b);
      for (i = 0; i < v.size(); i++) {
        d = v[i];
        e = a;
        while (e) {
          if (e % d == 0 && (e / d) % b != 0) {
            l = max(l, e / d);
            break;
          } else if (e % d == 0)
            e /= d;
        }
      }
      cout << l << "\n";
    }
  }
}
