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.
/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
import java.util.*;

public class A1444 {
    static long p, q;
    public static void main(String args[]) {
        Scanner sc=new Scanner(System.in);
        int test=sc.nextInt();
        while(test>0) {
            p=sc.nextLong();
            q=sc.nextLong();
            
            if(p%q==0) {
                long ans=1;
                for(long i=1; i*i<=q; i++) {
                    if(q%i==0) {
                        ans=Math.max(ans, fact(i));
                        ans=Math.max(ans, fact(q/i));
                    }
                }
                System.out.println(ans);
            }
            else {
                System.out.println(p);
            }
            
            test--;
        }
    }
    
    public static long fact(long i) {
        if(i==1)
            return 1;
        long a=p;
        while(a%q==0)
            a/=i;
        return a;
    }
}
