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;
const long long int inf = 1e15;
void solve() {
  long long int p, q;
  cin >> p >> q;
  if (p > q && p % q != 0) {
    cout << p << '\n';
  } else if (p >= q) {
    long long int val = q, mans = 0;
    vector<long long int> ans;
    for (long long int i = 2; i * i <= val; i++) {
      while (val % i == 0) {
        ans.push_back(i);
        val /= i;
      }
    }
    if (val > 1) ans.push_back(val);
    vector<pair<long long int, long long int> > res;
    long long int cnt = 1;
    for (long long int i = 1; i < ans.size(); i++) {
      if (ans[i] == ans[i - 1])
        cnt++;
      else {
        res.push_back({ans[i - 1], cnt});
        cnt = 1;
      }
    }
    res.push_back({ans[ans.size() - 1], cnt});
    for (long long int i = 0; i < res.size(); i++) {
      long long int tot = p;
      while (tot % (res[i].first) == 0) {
        tot /= res[i].first;
      }
      long long int mul = pow(res[i].first, res[i].second - 1);
      tot = tot * mul;
      mans = max(mans, tot);
    }
    cout << mans << '\n';
  } else if (p < q) {
    cout << p << '\n';
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
}
