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.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.io.*;
 
public class Division {
	
	private static class MyScanner {
	    BufferedReader br;
	    StringTokenizer st;

	    public MyScanner() {
	       br = new BufferedReader(new InputStreamReader(System.in));
	    }

	    String next() {
	        while (st == null || !st.hasMoreElements()) {
	            try {
	                st = new StringTokenizer(br.readLine());
	            } catch (IOException e) {
	                e.printStackTrace();
	            }
	        }
	        return st.nextToken();
	    }

	    int nextInt() {
	        return Integer.parseInt(next());
	    }

	    long nextLong() {
	        return Long.parseLong(next());
	    }

	    double nextDouble() {
	        return Double.parseDouble(next());
	    }

	    String nextLine(){
	        String str = "";
		  try {
		     str = br.readLine();
		  } catch (IOException e) {
		     e.printStackTrace();
		  }
		  return str;
	    }

	 }
	
	

	public static long solution(long p, long q)
			
	{
	   // TAKING INDIVIDUAL DIVISORS OF P AND CHECKING DIVISIBILTY BY Q WOULD BE O(ROOT P) IN TIME,
		// TO SOLVE FASTER, WE LOOK AT THE PRIME FACTORISATION OF Q AND P AND DIVIDE P BY PRIME FACTORS OF Q, TILL Q HAS SOME FACTOR THAT IS 
		// NOT PRESENT IN P, THIS WOULD BE A POSSIBLE VALUE FOR X
		// if p%q!=0, then answer is p
		// But if p%q==0 then, if p = 2^x * 3 ^y.... and q = 2^a * 3 ^b.... => x>=a, y>=b...
		// so if we divide p continuously by 2, say by 2^(x-a), we get p/2^(x-a) = 2^a * 3^y...
		// now, if we divide by 2 one more time, we get p/2^(x-a+1) = 2^(a-1) * 3^y..., i.e q has an addition factor 2 that is not in p/2^(x-a+1)
		// which implies p/2^(x-a+1) is not divisible by q, so we apply this logic for diff primes
		// (continuously divide p by the prime factors of q till it is no longer divisible by q and return the maximum of such values obtained)
		
		if(p%q!=0)
			return p;
		
		long max = 1;
		
		long s = q;
		for(int i = 2; i*i <= s; i++)
		{
			if(s%i==0)
			{
			    while(s%i==0)
				    s/=i;
   			    long x = p;
				while(x%q==0)
					x/=i;
				if(x>max)
					max = x;
			}
		}
		
		if(s>1)
		{
			while(p%q==0)
				p/=s;
			if(p>max)
				max = p;
		}
		
		return max;
			
	}
        
private static PrintWriter out = new PrintWriter(System.out);

public static void main (String[] args)
{
	MyScanner s =  new MyScanner();
     
    int t = s.nextInt();
    
    for(int j = 0; j<t ; j++)
    {
    	long p = s.nextLong();
    	long q = s.nextLong();

    	out.println(solution(p,q));
    }
    
    out.flush();
    out.close();
    
}
}