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.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;

public class Division {

    static long max(long a, long b) {
        if (a > b) {
            return a;
        }
        return b;
    }

    public static void main(String args[]) {
        Scanner scan = new Scanner(System.in);
        long t = scan.nextInt();
        k:
        while (t-- > 0) {
            long x = scan.nextLong();
            long y = scan.nextInt();
            if (x % y != 0) {
                System.out.println(x);
                continue k;
            }
            HashMap<Long, Long> h = new HashMap<Long, Long>();
            for (long i = 2; i <= Math.sqrt(y); i++) {
                long p = 0;
                if (y % i == 0) {
                    while (y % i == 0) {
                        y /= i;
                        ++p;
                    }
                    h.put(i, p);
                }
            }
            if (y >= 2) {
                h.put(y, (long) 1);
            }
            long res = 0;
            for (long k : h.keySet()) {
                long p = h.get(k);
                long c = x;
              
                while (c % k == 0) {
                    c /= k;
                }
                for (long i = 0; i < p - 1; i++) {
                    c *= k;
                }
                res = max(c, res);
            }
            System.out.println(res);
        }

    }

}
