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 Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int tz =sc.nextInt();
        for(int j=0;j<tz;j++) {
            long p=sc.nextLong();
            long q=sc.nextLong();
            if(p%q != 0){
                System.out.println(p);
            }
            else{
                ArrayList<Long> pf = new ArrayList<Long>();
                primeFactors(q,pf);

                long ans=0;
                for(long x:pf){
                    long t =p;
                    while(t%q==0){
                        t/=x;
                    }
                    ans=Math.max(ans,t);
                }

                System.out.println(ans);

            }
        }
    }

    private static void primeFactors(long n, ArrayList<Long> arr) {
        while (n % 2 == 0)
        {
            arr.add((long) 2);
            n = n/2;
        }

        for (int i = 3; i <= Math.sqrt(n); i = i + 2)
        {
            while (n % i == 0)
            {
                arr.add((long) i);
                n = n/i;
            }
        }

        if (n > 2)
            arr.add(n);
    }
}
