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.io.*;
import java.util.*;

public class Main {


    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner();
        PrintWriter out = new PrintWriter(System.out);
        int tc = sc.nextInt();
        loop:
        while (tc-- > 0) {
            long p = sc.nextLong();
            int q = sc.nextInt();
            int qq = q;
            long ans = p % q == 0 ? 1 : p;
            for (int i = 2; 1l * i * i <= q; i++) {
                if (q % i != 0) continue;
                while (q % i == 0)
                    q /= i;
                long pp = p;
                while (pp % qq == 0) pp /= i;
                ans = Math.max(ans, pp);
            }
            if (q != 1) {
                while (p % qq == 0) p /= q;
                ans = Math.max(ans, p);
            }
            out.println(ans);
        }
        out.flush();
        out.close();
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public Scanner(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(s));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public int[] nextIntArray(int n) throws IOException {
            int[] ans = new int[n];
            for (int i = 0; i < n; i++)
                ans[i] = nextInt();
            return ans;
        }

        public Integer[] nextIntegerArray(int n) throws IOException {
            Integer[] ans = new Integer[n];
            for (int i = 0; i < n; i++)
                ans[i] = nextInt();
            return ans;
        }

        public char nextChar() throws IOException {
            return next().charAt(0);
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        public boolean ready() throws IOException {
            return br.ready();
        }


    }
}