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.*;

public class A {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        StringBuilder sb = new StringBuilder();
        int testCases = sc.nextInt();

        for (int t = 0; t < testCases; t++) {
            long p = sc.nextLong();
            long q = sc.nextLong();

            if (p < q || p % q != 0) {
                sb.append(p).append("\n");
                continue;
            }

            Set<Integer> factors = new HashSet<>();

            long qc = q;

            while (qc % 2 == 0) {
                qc = qc / 2;
                factors.add(2);
            }

            for (int i = 3; i * i <= qc; i = i + 2) {
                while (qc % i == 0) {
                    factors.add(i);
                    qc = qc / i;
                }
            }

            if (qc > 2) {
                factors.add((int) qc);
            }

            long ans = Long.MIN_VALUE;

            for (int fact : factors) {

                long p_copy = p;
                long q_copy = q;

                while (p_copy % q_copy == 0) {
                    p_copy = p_copy / fact;
                }

                ans = Math.max(ans, p_copy);
            }

            sb.append(ans).append("\n");

        }

        System.out.print(sb);
    }
}