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.*;
public class division
{
    public static Vector<Long> fun(long a)
    {
        Vector<Long>arr=new Vector<Long>();
        arr.addElement(a);
        for(long i=2;i*i<=a;i++)
        {
            if(a%i==0)
            {
                arr.addElement(i);
                if(i!=a/i)
                {
                    arr.addElement(a/i);
                }
            }
        }
        
        return arr;
    }
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int t;
        t=sc.nextInt();
        while(t-->0)
        {
            long p,q;
            p=sc.nextLong();
            q=sc.nextLong();
            if(p%q!=0)
            {
                System.out.println(p);
            }
            else
            {
                Vector<Long>arr=fun(q);
                long ans=1;
                for(int i=0;i<arr.size();i++)
                {

                
                
                    long v=arr.get(i);
                    long tmpp=p;
                    while(tmpp%v==0)
                    {
                        tmpp/=v;
                         if(tmpp%q!=0)
                       {
                        ans=Math.max(ans,tmpp);
                        break;
                       }
                    }
                   
                
                }
               System.out.println(ans);
            }
        }
    }

}