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() {
  long long int t;
  cin >> t;
  for (long long int i = 0; i < t; i++) {
    long long int a, b;
    cin >> a >> b;
    vector<pair<long long int, long long int> > v;
    long long int cnt2 = 0, cnt = 0, bb = b;
    if (b % 2 == 0) {
      while (b % 2 == 0) {
        cnt2++;
        b >>= 1;
      }
      v.push_back(make_pair(2, cnt2));
    }
    for (long long int i = 3; i * i <= b; i++) {
      cnt = 0;
      if (b % i == 0) {
        while (b % i == 0) {
          cnt++;
          b = b / i;
        }
        v.push_back(make_pair(i, cnt));
      }
    }
    if (b != 1) v.push_back(make_pair(b, 1));
    long long int aa = a, ans = 1;
    for (long long int i = 0; i < v.size(); i++) {
      int cnt1 = 0;
      long long int x = v[i].first, y = v[i].second;
      while (a % x == 0) {
        a /= x;
        cnt1++;
      }
      cnt1 = cnt1 - v[i].second;
      cnt1++;
      long long int k2 = 1;
      for (long long int j = 1; j <= cnt1; j++) {
        k2 *= v[i].first;
      }
      ans = max(ans, aa / k2);
    }
    cout << ans << '\n';
  }
}
