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 A {
    static PrintWriter ww = new PrintWriter(System.out);
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int tt = sc.nextInt();
        while(tt -- >  0){
            long p = sc.nextLong();
            long q = sc.nextLong();
            
            if(p%q != 0)ww.println(p);
            else{
                ArrayList<Long> arr = new ArrayList<>();
                long temp = q;
                for(long i = 2 ; i*i <= q ; i++){
                    if(temp%i == 0){
                        while(temp % i == 0){
                            temp /= i;
                        }
                        arr.add(i);
                    }
                }
                if(temp != 1){
                    arr.add(temp);
                }
                long ans = p , res = p;
                for(long i : arr){
                    int ca = 0 , cb = 0;
                    long ta = p , tb = q;
                    while(ta % i == 0){
                        ta /= i;
                        ca++;
                    }
                    while(tb % i == 0){
                        tb /= i;
                        cb++;
                    }
                    long min = 1;
                    while(ca >= cb){
                        min *= i;
                        ca--;
                    }
                    res = Math.min(res, min);
                }
                ww.println(ans/res);

            }
        }
        ww.close();
    }    
}
