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.
//make sure to make new file!
import java.io.*;
import java.util.*;

public class A680{
   
   public static void main(String[] args)throws IOException{
      BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter out = new PrintWriter(System.out);
      
      int t = Integer.parseInt(f.readLine());
      
      for(int qq = 1; qq <= t; qq++){
      
         StringTokenizer st = new StringTokenizer(f.readLine());
      
         long p = Long.parseLong(st.nextToken());
         long q = Long.parseLong(st.nextToken());
         
         if(p > q && p%q != 0) out.println(p);
         else if(p < q) out.println(p);
         else{
            //try different prime factors of q
            long answer = 1L;
            long x = q;
            long i = 2L;
            while(i*i <= x){
               if(x % i == 0){
                  long curp = p;
                  while(curp % q == 0){
                     curp/=i;
                  }
                  answer = Math.max(answer,curp);
               }
               while(x % i == 0) x/=i;
               i++;
            }
            
            if(x > 1){
               long curp = p;
                  while(curp % q == 0){
                     curp/=x;
                  }
                  answer = Math.max(answer,curp);
            }
            
            out.println(answer);
         }
      }
      
      
      
      
      out.close();
   }
   
      
}