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 {
    static class Pair{
        long p, q, x;
        Pair(long p, long q){
            this.p = p;
            this.q = q;
            this.x = p;
            this.findx();
        }
        void findx(){
            if(p < q || (p > q && p%q!=0)) return;
            long j=2L, ans=q;
            ArrayList<Long> a = new ArrayList<Long>();
            while(j*j<=q){
                if(ans%j==0){
                    a.add(j);
                    while(ans%j==0){
                        ans /= j;
                    }
                }
                j++;
            }
            if(ans>1){
                a.add(ans);
                ans = 1;
            }
            for(long l: a){
                long k = p;
                while(k%q==0){
                    k /= l;
                }
                ans = Math.max(ans, k);
            }
            this.x = ans;
        }
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.nextLine());
        Pair pairs[] = new Pair[n];
        for(int i=0;i<n;i++){
            String num[] = sc.nextLine().split(" ");
            pairs[i] = new Pair(Long.parseLong(num[0]),Long.parseLong(num[1]));
        }
        for(int i=0;i<n;i++){
            print(pairs[i].x);
        }
        sc.close();
    }
    public static void print(long s){
        System.out.println(s);
    }
}