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.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.Locale;
import java.util.Scanner;
import java.util.Set;

public class Main {

    public static void main(String[] args) throws Exception {
        try (BufferedInputStream in = new BufferedInputStream(System.in);
             PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out))) {

            Scanner sc = new Scanner(in).useLocale(Locale.US);

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

                Set<Long> primes = primes(p);
                long max = 0;
                for (Long prime : primes) {
                    long tmp = q;
                    while (tmp % p == 0) tmp /= prime;
                    max = Math.max(max, tmp);
                }

                out.println(max);
            }
        }
    }

    private static Set<Long> primes(long v) {
        Set<Long> primes = new HashSet<>();
        for (long i = 2; i * i <= v; i++) {
            while (v % i == 0) {
                primes.add(i);
                v /= i;
            }
        }
        if (v > 1) {
            primes.add(v);
        }
        return primes;
    }
}
