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.ArrayList;
import java.util.HashMap;
import java.util.Scanner;
import java.util.Set;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s = new Scanner(System.in);
		int tc = s.nextInt();
		for(int t = 0;t < tc;t++) {
			long p = s.nextLong();
			long q = s.nextLong();
			
			if(p % q != 0) {
				System.out.println(p);
				continue;
			}
			
			ArrayList<Integer> arr = new ArrayList<>();
			//boolean[] is_prime  = new boolean[(int)q + 1];
			HashMap<Integer , Integer> prime = new HashMap<>();
			HashMap<Integer , Integer> map = new HashMap<>();
			
//			for(int i = 2;i * i <= q;i++) { // O(sqrt(q))
//				if(!map.containsKey(i)) {
//					prime.put(i , 1);
//					for(int j = i * i;j <= q;j += i) {
//						map.put(j , 1);
//					}
//				}
//			}
			ArrayList<Long> result = primeFactors(q);
			long re = 0;
			for(int i = 0;i < result.size();i++) {
				long temp = p;
				
				while(temp % q == 0) {
					temp /= result.get(i);
				}
				re = Math.max(re, temp);
				
			}
			System.out.println(re);
//			for(int i = 2;i <= (int)Math.sqrt(q);i++) {
////				if(q%i == 0) {
////					if(!map.containsKey(i) && !map.containsKey((int)q/i)) {
////						prime.put(i , 1);
////						prime.put((int)q/i , 1);
////					}
////				}
////				else 
//				if(!map.containsKey(i)) prime.put(i , 1);
//			}
////			for(int i = 2;i <= q;i++) {
////				if(prime.containsKey(i)) {
////					arr.add(i);
////				}
////			}
//			Set<Integer> keySet = prime.keySet();
//			for(int i : keySet) System.out.print(i + " ");
//			System.out.println();
//			long ans = 1;
//			for(int i : keySet) {	
//				long temp = p;
//				
//				while(temp % i == 0) { // O(log2p)
//					temp /= i;
//					if(temp % q != 0) {
//						ans = Math.max(ans, temp);
//						break;
//					}
//				}
//			}
//			
//			System.out.println(ans);
			
		}
	}
	
	public static ArrayList<Long> primeFactors(long num) {
		ArrayList<Long> arr = new ArrayList<>();
		
		while(num % 2 == 0) {
			arr.add((long)2);
			num /= 2;
		}
		
		for(int i = 3;i * i <= num;i += 2) {
			while(num % i == 0) {
				arr.add((long)i);
				num /= i;
			}
		}
		
		if(num >= 2) arr.add(num);
		
		return arr;
	}

}
