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 Main{
    public static void main(String[] args) throws IOException{
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        for(int t = pint(in); t > 0; t--) {
        	StringTokenizer st = new StringTokenizer(in.readLine()); 
        	long p = Long.parseLong(st.nextToken()), q = pint(st);
        	if(p % q != 0) {
        		System.out.println(p);
        		continue;
        	}
        	long max = 1;
        	for(int i = 2; i <= 31623; i++) {
        		if(q % i != 0) {continue;}
        		long c = 1;
        		while(q % i == 0) {
        			c *= i;
        			q /= i;
        		}
        		long s = p;
        		while(s % c == 0) {
        			s /= i;
        		}
        		if(s > max) {max = s;}
        	}
        	if(q != 1) {
        		long s = p;
        		while(s % q == 0) {
        			s /= q;
        		}
        		if(s > max) {max = s;}
        	}
        	System.out.println(max);
        }
    }
    static int pint(BufferedReader in) throws IOException {return Integer.parseInt(in.readLine());}
    static int pint(StringTokenizer st) {return Integer.parseInt(st.nextToken());}
}