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 MOD = 1e9 + 7;
long long int power(long long int x, long long int y) {
  if (y == 0) return 1;
  long long int temp = power(x, y / 2);
  if (y % 2 == 0)
    return temp * temp;
  else
    return ((x * temp) * temp);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) {
    long long int p, q, i, j;
    cin >> p >> q;
    std::map<long long int, long long int> m;
    long long int q1 = q;
    for (i = 2; i * i <= q; i++) {
      if (q1 % i == 0) {
        long long int x = 0;
        while (q1 % i == 0) q1 /= i, x++;
        m[i] = x;
      }
    }
    if (q1 > 1) m[q1] = 1;
    std::map<long long int, long long int> fac;
    pair<long long int, long long int> z = {-1, -1};
    long long int mn = 1e18, f = 0;
    for (auto x : m) {
      long long int num = 0;
      while (p % x.first == 0) {
        num++;
        p /= x.first;
      }
      if (num >= x.second) {
        if (power(x.first, num - x.second + 1) < mn) {
          mn = power(x.first, num - x.second + 1);
          z = {x.first, x.second - 1};
        }
      } else
        f = 1;
      fac[x.first] = num;
    }
    long long int ans = 1;
    for (auto x : fac) {
      if (f == 0 && z.first == x.first) {
        ans *= power(z.first, z.second);
      } else
        ans *= power(x.first, x.second);
    }
    ans *= p;
    cout << ans << "\n";
  }
  cerr << "Time Taken:" << (float)clock() / CLOCKS_PER_SEC * 1000 << "ms"
       << "\n";
  return 0;
}
