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 Solution{
	static PrintWriter out=new PrintWriter(System.out);
	public static void main (String[] args) throws IOException{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		
		String[] input=br.readLine().trim().split(" ");
		
		int numTestCases=Integer.parseInt(input[0]);
		while(numTestCases-->0){
		    input=br.readLine().trim().split(" ");
		    long p = Long.parseLong(input[0]);
		    long q = Long.parseLong(input[1]);
		    
		    out.println(getX(p, q));
		}
		
		out.flush();
		out.close();
	}
	public static long getX(long p, long q)
	{
	    if(p % q != 0)
	    {
	        return p;
	    }
	    long ans = 1;
	    long temp2 = q;
	    for(int i = 2; i <= Math.sqrt(q); i++){
	        if(q % i == 0){
    	        long temp = p;
    	        while(temp % i == 0 && temp % temp2 == 0){
    	            temp /= i;
    	        }
    	        while(q % i == 0)
    	        {
    	            q /= i;
    	        }
    	        ans = Math.max(ans, temp);
	        }
	    }
	    if(q > 1)
	    {
	        long temp = p;
	        while(temp % temp2 == 0){
	            temp /= q;
	        }
	        ans = Math.max(ans, temp);
	    }
	    
	    return ans;
	}
}