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.Scanner;

public class Main {
	public long calc(long p, long q, long i) {
		long l;
		l = i;
		while(q%(l*i) == 0) {
			l *= i;
		}
	    
    	while(p%i == 0) {
    		p = p/i ;
    	}
    	return p*(l/i);
	}
	public static void main(String [] args) {
		Scanner sc = new Scanner(System.in) ;
		Main mn = new Main();
		int t = sc.nextInt() ;
		long ii,j, p, q, i, k, l;
		for(ii=0;ii<t;ii++){
		    p=sc.nextLong() ;
		    q=sc.nextLong() ;
		    
		    if(p%q != 0) {
		    	System.out.println(p);
		    	continue;
		    }
		    j=(int)Math.sqrt(q);
		    i = 1 ;
		    k = 1 ;
		    while(i<=j) {
		    	if(q%i == 0) {
		    		if( i!= 1) k = Math.max(k, mn.calc(p, q, i));
		    		k = Math.max(k, mn.calc(p, q, q/i));
		    	}
		    	i++ ;
		    }
		    System.out.println(k);
		 }
		sc.close();
	}
}
