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;

/**
 * Accomplished using the EduTools plugin by JetBrains https://plugins.jetbrains.com/plugin/10081-edutools
 */

public class Main {
    private static TreeSet<Long> primeFactors(long number) {
        TreeSet<Long> list = new TreeSet<>();
        long factor = 2;
        long tmp = number;
        while (tmp > 1 && factor <= Math.min(tmp, Math.sqrt(number))) {
            if (tmp % factor == 0) {
                tmp /= factor;
                list.add(factor);
            } else {
                factor++;
            }
        }
        if (tmp > 1)
            list.add(tmp);
        return list;
    }

    public static void main(String[] args) {


        // Write your solution here
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        for (int i = 0; i < n; i++) {
            long a = sc.nextLong();
            long b = sc.nextLong();


            //make it more efficient
            //search for divisors bottom up
            /*
            for (int j = 1; j <= Math.sqrt(a); j++) {
                if (a % j == 0) {
                    long x = a / j;
                    if (x % b != 0) {
                        System.out.println(x);
                        break;
                    }
                }
            }
             */


            //x is a, if b is not a divisor of a
            if (a % b != 0)
                System.out.println(a);
            else {
                //find all prime factors of b
                TreeSet<Long> factors = primeFactors(b);

                //find the largest remainder, when one prime factor of b is divided out completely
                long max = 1;
                for (Long factor : factors) {
                    long tmp = a;
                    while (tmp % factor == 0) {
                        tmp /= factor;
                        if (tmp % b != 0) {
                            max = Math.max(tmp, max);
                            break;
                        }
                    }
                }
                System.out.println(max);

            }
        }
    }
}