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;
long long int sumOfdigit(long long int n) {
  long long int sum = 0;
  while (n) {
    long long int k = n % 10;
    sum += k;
    n /= 10;
  }
  return sum;
}
int arr[100];
int tree[100 * 3];
void segment(int node, int b, int e) {
  if (b == e) {
    tree[node] = arr[b];
  }
  int l = node * 2;
  int r = node * 2 + 1;
  int mid = (l + r) / 2;
  segment(l, b, mid);
  segment(r, mid + 1, e);
  tree[node] = tree[l] + tree[r];
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long int p, q;
    cin >> p >> q;
    vector<long long int> v;
    long long int qq = q;
    for (int i = 2; i * i <= qq; i++) {
      if (q % i == 0) {
        v.push_back(i);
        while (qq % i == 0) {
          qq /= i;
        }
      }
    }
    if (qq > 1) {
      v.push_back(qq);
    }
    long long int ans = 1;
    for (int i = 0; i < v.size(); i++) {
      long long int pp = p;
      while (pp % q == 0) {
        pp /= v[i];
      }
      ans = max(ans, pp);
    }
    cout << ans << '\n';
  }
  return 0;
}
