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 calculate(long long int p, long long int q) {
  long long int modx = 998244353, expo;
  expo = modx - 2;
  while (expo) {
    {
      if (expo & 1) p = (p * q) % modx;
    }
    q = (q * q) % modx;
    expo >>= 1;
  }
  return p % modx;
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int a, b;
    cin >> a >> b;
    if (b > a)
      cout << a << "\n";
    else if (a % b != 0)
      cout << a << "\n";
    else {
      long long int y = b, x = sqrt(b);
      vector<long long int> z;
      for (long long int i = 2; i <= x; i++) {
        while (y % i == 0) {
          z.push_back(i);
          y = y / i;
        }
      }
      if (y != 1) z.push_back(y);
      long long int c = 0;
      x = z.size();
      y = a / b;
      for (long long int i = 0; i < x; i++) {
        long long int u = b / z[i], v = y;
        while (v % z[i] == 0) v = v / z[i];
        c = max(c, u * v);
      }
      cout << c << "\n";
    }
  }
  return 0;
}
