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.
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		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){
			System.out.println(a);
		}
		else {
		ArrayList<Long> al=new ArrayList<>();
		long z=(long)Math.sqrt(b),i;
			for(i=2;i<=z;i++){
				if(b%i==0){
					al.add(i);
					if(i!=(b/i))
				al.add(b/i);
				}
			}
			Collections.sort(al);
			al.add(b);
			long x=a/b;
			long ans=1;
			for(i=al.size()-1;i>=0;i--){
				long an=1,res=x;
					while(res%al.get((int)i)==0)
					res/=al.get((int)i);
					an=res*(b/al.get((int)i));
			
					if(an%b>0)
					ans=Math.max(ans,an);
			}
			System.out.println(ans);
		    
		}
		}
	}
}
