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.*;
import java.util.Collections;

 public class S96{
     static ArrayList<Long> ar=new ArrayList<>();
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0){
	long a=sc.nextLong();
	long b=sc.nextLong();
	if(a%b!=0 || (a>b && a%b!=0)){
	    System.out.println(a);
	   
	}
	else{
	     isprime(b);long ans=0;
	     Collections.sort(ar,Collections.reverseOrder());
	     for(long aa:ar){
	         long an=a;
	         while(an%aa==0){
	             an=an/aa;
	       
	             
	             if((an)%b!=0)
	             {
	             ans=Math.max(an,ans);}
	         }
	     
	     }
	     System.out.println(ans);
	
		}
		
		
		
		
		
		
		}
	}
		
	
	
static void isprime(long n){
    ar.add(n);
    
    for(long i=2;i*i<=n;i++){
        if(n%i==0){
          
         
            ar.add(i);
            if(i!=n/i){
                ar.add(n/i);
            }
        }
        
        
    }
    
    
}
}