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.*;

public class Solution {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        StringBuilder stringBuilder = new StringBuilder();
        int t = scanner.nextInt();
        for (int i = 0; i < t; ++i) {
            long p = scanner.nextLong();
            long q = scanner.nextLong();
            if (p % q != 0) {
                stringBuilder.append(p).append('\n');
                continue;
            }
            long z = 2;
            long cq = q;
            long ans = 1;
            while (z * z <= cq) {
                long jihad = 1;
                while (cq % z == 0) {
                    jihad *= z;
                    cq /= z;
                }
                if (jihad != 1) {
                    long puk = p;
                    while (puk % jihad == 0) puk /= jihad;
                    while (puk % z == 0) puk /= z;
                    puk *= jihad;
                    puk /= z;
                    ans = Math.max(ans, puk);
                }
                ++z;
            }
            if (cq > 1) {
                long puk = p;
                while (puk % cq == 0) puk /= cq;
                ans = Math.max(ans, puk);
            }
            stringBuilder.append(ans).append('\n');
        }
        System.out.println(stringBuilder);

    }

}
