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;
void kehsihba() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void solve() {
  long long a, b;
  cin >> a >> b;
  if (a % b != 0) {
    cout << a;
    cout << '\n';
    ;
    return;
  }
  set<long long> s;
  for (long long i = 1; i * i <= b; i++) {
    if (b % i == 0) {
      s.insert(i), s.insert(b / i);
    }
  }
  long long ans = -1;
  for (auto lol : s) {
    if (lol == *s.begin()) {
      continue;
    }
    long long x = a / lol;
    while (x % b == 0) {
      x = x / lol;
    }
    ans = max(ans, x);
  }
  cout << ans;
  cout << '\n';
  ;
}
int32_t main() {
  kehsihba();
  long long t = 1;
  cin >> t;
  for (long long k = 1; k <= t; k++) {
    solve();
  }
}
