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.
//package division;

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

public class division {
	
	public static boolean isPrime(long val) {
		for(int i = 2; i <= (int) Math.sqrt(val); i++) {
			//System.out.println("YES");
			if(val % i == 0) {
				return false;
			}
		}
		return true;
	}
	
	public static void main(String[] args) throws IOException {
		BufferedReader fin = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(fin.readLine());
		StringBuilder fout = new StringBuilder();
		while(t-- > 0) {
			StringTokenizer st = new StringTokenizer(fin.readLine());
			long q = Long.parseLong(st.nextToken());
			long k = Long.parseLong(st.nextToken());
			if(q % k != 0) {
				fout.append(q).append("\n");
			}
			else {
				long ans = 0;
				if(isPrime(k)) {
					while(q % k == 0) {
						q /= k;
					}
					ans = q;
				}
				else {
					long multiple = q / k;
					long multiple2 = multiple;
					long k2 = k;
					for(int i = 2; i <= (int) Math.sqrt(k2); i++) {
						//System.out.println(Math.sqrt(k2));
						if(k2 % i == 0) {
							//System.out.println(i + " ----");
							k = k2;
							multiple = multiple2;
							int next = (int) (k / i);
							int qAmt = 0;
							long temp = q;
							while(k % i == 0) {
								k /= i;
							}
							while(multiple % i == 0) {
								//System.out.println("R");
								multiple /= i;
								qAmt ++;
							}
							for(int j = 0; j < qAmt + 1; j++) {
								//System.out.println(j);
								temp /= i;
							}
							ans = Math.max(temp, ans);
							int old = i;
							k = k2;
							i = next;
							multiple = multiple2;
							qAmt = 0;
							temp = q;
							while(k % i == 0) {
								k /= i;
							}
							while(multiple % i == 0) {
								//System.out.println("R");
								multiple /= i;
								qAmt ++;
							}
							for(int j = 0; j < qAmt + 1; j++) {
								//System.out.println(j);
								temp /= i;
							}
							ans = Math.max(temp, ans);
							//System.out.println(old + " " + next);
							i = old;
						}
					}
				}
				fout.append(ans).append("\n");
			}
		}
		System.out.print(fout);
	}
}
