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 Problem1444a {
    static long pi;
    static long qi;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-->0) {
            pi = sc.nextLong();
            qi = sc.nextLong();
            long tmpp = pi;
            long tmpq = qi;
            long ans = 0;
            for (int i = 2; i <= (long) (Math.sqrt(tmpq)); i++) {
                tmpp = pi;
                while (tmpq % i == 0) {
                    if (tmpp % qi != 0) {
                        ans = Math.max(ans, tmpp);
                    }
                    else if (tmpp % i != 0) {
                    }
                    else {
                        tmpp /= i;
                        continue;
                    }
                    tmpq /= i;
                }
            }
            if (tmpq >= 2) {
                long x = pi;
                while (true) {
                    if (x % qi != 0) {
                        ans = Math.max(ans, x);
                        break;
                    }
                    if (x % tmpq != 0) {
                        break;
                    }
                    else x /= tmpq;
                }
            }
            System.out.println(ans);
        }
    }
}
