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.
import java.util.Scanner;
import java.util.Vector;

public class C {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while (t-- >0){
            long p, q;
            p=sc.nextLong();
            q=sc.nextLong();
            Vector<Long>div=new Vector<>();
            for (long i = 2; i * i <= q; ++i) {
                if (q % i == 0) {
                    div.add(i);
                    if (q / i != i)
                        div.add(q / i);
                }
            }
            div.add(q);
            long ans = 0;
            for (long a : div){
                long temp = p;
                while (temp % a == 0){
                    if (temp % q != 0 && p % temp == 0)
                    ans = Math.max(ans, temp);
                    temp /= a;
                }
                ans = Math.max(ans, temp);
            }
            System.out.println(ans);
        }
    }
}
