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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class A {

    public static void main(String[] args) {
        FastScanner fs=new FastScanner();
        try {

            int T = fs.nextInt();
            int ans=0;

            for (int tt = 0; tt < T; tt++) {
                 Long p = fs.nextLong();
                 int q=fs.nextInt();




                if(p%q==0){
                    HashSet<Integer> factors = new HashSet<Integer>();
                    long Q=q;
                    while (q%2==0)
                    {
                        factors.add(2);
                        q /= 2;
                    }
                    for (int i = 3; i <= Math.sqrt(q); i+= 2)
                    {
                        while (q%i == 0)
                        {
                            factors.add(i);
                            q /= i;
                        }
                    }
                   
                    if (q > 2)
                    factors.add(q);



                     long x=1;
                     for(int  i:factors){



                             long newp = p;
                             while ((newp % i == 0)) {
                                 newp /= i;

                                 if (newp % Q!= 0) {
                                     x = Math.max(newp, x);
                                     break;

                                 }
                             }


                             }





                     System.out.println(x);
                 }




                 else {
                     System.out.println(p);

                 }






            }
        }
        catch(Exception e){
            return;}



    }



    static void sort(int[] a) {
        ArrayList<Integer> l=new ArrayList<>();
        for (int i:a) l.add(i);
        Collections.sort(l,Collections.reverseOrder());

        for (int i=0; i<a.length; i++) a[i]=l.get(i);
    }

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }


}
