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>
#pragma GCC optimize("-O2")
using namespace std;
const int LIM = 2e5 + 5, MOD = 1e9 + 7;
const long double EPS = 1e-9;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tests = 1;
  cin >> tests;
  while (tests--) {
    long long p, q;
    long long op;
    cin >> p >> q;
    op = p;
    long long mi = p;
    for (int i = 2; i < 1e5; ++i) {
      if ((q % i) == 0) {
        long long c = 1;
        long long num = 0;
        while ((q % i) == 0) {
          num++;
          q /= i;
        }
        long long num2 = 0;
        while ((p % i) == 0) {
          num2++;
          p /= i;
        }
        for (int j = 0; j < num2 - num + 1; ++j) {
          c *= i;
        }
        mi = min(mi, c);
      }
    }
    if (q > 1) {
      long long num1 = 1;
      long long num2 = 0;
      while ((p % q) == 0) {
        p /= q;
        num2++;
      }
      long long c = 1;
      for (int j = 0; j < num2 - num1 + 1; ++j) {
        c *= q;
      }
      mi = min(mi, c);
    }
    cout << op / mi << '\n';
  }
  return 0;
}
