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.TreeSet;

public class p1445C {
    public static void main(String[] args) {
        Scanner sr = new Scanner(System.in);
        int t = sr.nextInt();
        for (int i = 0; i < t; i++) {
            long p = sr.nextLong();
            long q = sr.nextLong();
            if (p % q != 0) {
                System.out.println(p);
            }
            else {
                long max = 0;
                TreeSet<Long> primes = new TreeSet<>();
                long temp_q = q;
                for (long j = 2; j * j <= temp_q; j++) {
                    while (temp_q % j == 0) {
                        primes.add(j);
                        temp_q /= j;
                    }
                }
                if (temp_q > 1) {
                    primes.add(temp_q);
                }
                for (long j: primes) {
                    long temp_p = p;
                    while (temp_p % q == 0) {
                        temp_p /= j;
                    }
                    max = Math.max(max, temp_p);
                }
                System.out.println(max);
            }
        }
    }
}