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.io.*;
public class q3{
    public static void main(String args[]) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int test = Integer.parseInt(br.readLine());
        while(test-->0){
            StringTokenizer st = new StringTokenizer(br.readLine());
            long p = Long.parseLong(st.nextToken());
            long q = Long.parseLong(st.nextToken());
            if(p%q!=0){
                System.out.println(p);
            }
            else{
                HashSet<Long> set = new HashSet<>();
                long temp = q;
                long max = -1;
                while(q%2==0){
                    q= q/2;
                    set.add((long)2);
                }
                for(long i=3;i<=Math.sqrt(temp);i+=2){
                    while(q%i==0){
                        q = q/i;
                        set.add(i);
                    }
                }
                if(q>2){
                    set.add(q);
                }
                for(long i:set){
                    //System.out.println(i+" ");
                    long tt = p;
                    while(tt%i==0){
                        tt = tt/i;
                        if(tt%temp!=0){
                          max =   Math.max(max,tt);
                        }
                    }
                }
                System.out.println(max);

            }
        }
    }
}