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.math.BigInteger;
import java.util.Scanner;


public class C_2 {
    public static void main(String[] args) {
        BigInteger t,n,m,ali;
        Scanner scan = new Scanner(System.in);
        t = scan.nextBigInteger();
        while(t.compareTo(BigInteger.ZERO) != 0) {
            t = t.subtract(BigInteger.ONE);
            n = scan.nextBigInteger();
            m = scan.nextBigInteger();
            if (n.compareTo(m) < 0) {
                System.out.println(n);
                continue;
            }
            ali = BigInteger.ZERO;
            for (BigInteger j = BigInteger.ONE; (j.multiply(j)).compareTo(m) <= 0; j = j.add(BigInteger.ONE)) {
                if ((m.mod(j)).compareTo(BigInteger.ZERO) != 0)
                    continue;
                if (j.compareTo(BigInteger.ONE) != 0) {
                    BigInteger k = n;
                    while ((k.mod(m)).compareTo(BigInteger.ZERO) == 0)
                        k = k.divide(j);
                    if (ali.compareTo(k) < 0)
                        ali = k;
                }
                BigInteger k = n;
                while ((k.mod(m)).compareTo(BigInteger.ZERO) == 0) {
                    BigInteger a = m.divide(j);
                    k = k.divide(a);
                }
                if (ali.compareTo(k) < 0)
                    ali = k;
            }
            System.out.println(ali);
        }
    }
}