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 _1445_C {
	public static void main(String[] args) throws Exception {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		int t = Integer.parseInt(in.readLine());
		for(int i = 0; i < t; i++) {
			StringTokenizer line = new StringTokenizer(in.readLine());
			long pi = Long.parseLong(line.nextToken());
			long qi = Long.parseLong(line.nextToken());
			ArrayList<Long> factors = primeFactorize(qi);
			long res = 0;
			for(int j = 0; j < factors.size(); j++) {
				long curfactor = factors.get(j);
				long curdiv = pi;
				while(curdiv % qi == 0) {
					curdiv /= curfactor;
				}
				res = Math.max(res, curdiv);
			}
			out.println(res);
		}
		in.close();
		out.close();
	}
	static ArrayList<Long> primeFactorize(long num) {
		if(num == 0) return new ArrayList<Long>();
		ArrayList<Long> res = new ArrayList<Long>();
		while(num % 2 == 0) {
			res.add((long)2);
			num /= 2;
		}
		for(int i = 3; i * i <= num; i+=2) {
			while(num % i == 0) {
				res.add((long)i);
				num /= i;
			}
		}
		if(num >= 2) {
			res.add(num);
		}
		return res;
	}
}
