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 long long int mod = 1000000007;
const long long int INF64 = mod * mod + 1;
void solve() {
  long long int p, q;
  cin >> p >> q;
  if (p % q) {
    cout << p << '\n';
    return;
  }
  long long int ans = 1;
  for (long long int i = 2; i * i <= q; i++) {
    if (q % i == 0) {
      long long int b = 0;
      long long int temp1 = p, temp2 = q;
      while (p % i == 0) p /= i;
      while (q % i == 0) {
        q /= i;
        b++;
      }
      b--;
      while (b--) p *= i;
      ans = max(ans, p);
      p = temp1, q = temp2;
    }
    if (q % i == 0) {
      long long int b = 0;
      long long int temp1 = p, temp2 = q;
      while (p % (temp2 / i) == 0) p /= (temp2 / i);
      while (q % (temp2 / i) == 0) {
        q /= (temp2 / i);
        b++;
      }
      b--;
      while (b--) p *= (temp2 / i);
      ans = max(ans, p);
      p = temp1, q = temp2;
    }
  }
  if (ans == 1) {
    while (p % q == 0) p /= q;
    ans = max(ans, p);
  }
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int T = 1;
  cin >> T;
  for (long long int i = 1; i <= T; i++) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}
