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;
int main() {
  int n;
  int t;
  cin >> t;
  for (int qwerty = 0; qwerty < t; ++qwerty) {
    int64_t p, q, pp;
    cin >> p >> q;
    pp = p;
    int64_t ans = 1e18;
    for (int i = 2; i * i <= q; ++i) {
      int cnt1 = 0;
      int cnt2 = 0;
      while (p % i == 0) {
        p /= i;
        cnt1++;
      }
      while (q % i == 0) {
        q /= i;
        cnt2++;
      }
      if (cnt1 < cnt2) ans = 1;
      if (cnt1 != 0 && cnt2 != 0) {
        int cnt = max(0, cnt1 - cnt2 + 1);
        int64_t a = 1;
        for (int j = 0; j < cnt; ++j) a *= i;
        ans = min(ans, a);
      }
    }
    int cnt1 = 0;
    int cnt2 = 0;
    if (q > 1) {
      int qq = q;
      while (p % q == 0) {
        p /= q;
        cnt1++;
      }
      cnt2 = 1;
      if (cnt1 < cnt2) {
        ans = 1;
      } else {
        int cnt = max(0, cnt1 - cnt2 + 1);
        int64_t a = 1;
        for (int j = 0; j < cnt; ++j) a *= q;
        ans = min(ans, a);
      }
    }
    cout << pp / ans << endl;
  }
}
