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.io.BufferedOutputStream;
import java.io.PrintStream;
import java.util.Scanner;


public class Solution {

    private static boolean TESTS = true;

    private final Scanner in;
    private final PrintStream out;

    public Solution(final Scanner in, final PrintStream out) {
        this.in = in;
        this.out = out;
    }


    public void solve() {
        for (int tests = TESTS ? in.nextInt() : 1; tests > 0; tests--) {
            final long p = in.nextLong();
            final long q = in.nextLong();

            long x = 1;

            if (q > p || p % q != 0) {
                x = p;
            } else if (p % q == 0) {
                long a = q;

                for (long i = 2; i * i <= a; i++) {
                    if (a % i == 0) {
                        while (a % i == 0) a /= i;

                        long b = p;
                        while (b % q == 0) b /= i;

                        x = Math.max(x, b);
                    }
                }

                if(a > 1) {
                    long b = p;
                    while (b % q == 0) b /= a;

                    x = Math.max(x, b);
                }

            }

            out.println(x);
        }
    }


    public static void main(final String[] args) {
        final Scanner in = new Scanner(System.in);
        final PrintStream out = new PrintStream(new BufferedOutputStream(System.out));

        try {
            new Solution(in, out).solve();
        } finally {
            out.flush();
        }
    }

}
