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.*; 
import java.math.*; 

public class Main { 
	public static HashMap<Integer,Integer> primeFactors(long n) 
    { 
		HashMap<Integer,Integer> ans=new HashMap<>();
        int temp=0;
        while (n%2==0) 
        { 
            n /= 2; 
            temp+=1;
        } 
        if(temp>0) {
        	ans.put(2,temp);
        }
        
//        System.out.println(Math.sqrt(n));
        for (int i = 3; i <= Math.sqrt(n); i+= 2) 
        { 
            temp=0;
            
            while (n%i == 0) 
            { 
                temp++; 
                n /= i; 
            } 
            if(temp>0) {
            	ans.put(i,temp);
            }
//            System.out.println(n+" "+i);
        } 
        if (n > 2) {
        	ans.put((int)n, 1);
        }
        return ans;	
    } 
	public static void main(String[] args) throws IOException 
	{ 
		FastScanner f = new FastScanner(); 
		int t=f.nextInt();
//		int t=1;
		while(t-->0){
			long n=f.nextLong();
			long m=f.nextLong();
			if(n%m!=0) {
				System.out.println(n);
			}
			else {
				long ans=-1;
				HashMap<Integer,Integer> a2=primeFactors(m);
//				System.out.println(a1);
//				System.out.println(a2);
				for(int i:a2.keySet()) {
					long temp=n;
					int num=0;
					while(temp%m==0) {
						temp/=i;
					}
					ans=Math.max(ans,temp);
				}
				System.out.println(ans);
			}
		}
	} 
	static void sort(int [] a) {
        ArrayList<Integer> q = new ArrayList<>();
        for (int i: a) q.add(i);
        Collections.sort(q);
        for (int i = 0; i < a.length; i++) a[i] = q.get(i);
    }
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}
} 