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.*;
import java.io.*;

public class A1444 {

	static void getDiv(int idx, long cur) {
		if (idx == factors.size()) {
			div.add(cur);
			return;
		}
		long[] x = factors.get(idx);
		long mult = 1;
		for (int i = 0; i <= x[1]; i++) {
			getDiv(idx + 1, cur * mult);
			mult *= x[0];
		}
	}

	static ArrayList<long[]> factors;
	static ArrayList<Long> div;

	static void getFactors(long x) {
		factors = new ArrayList<long[]>();
		div = new ArrayList<Long>();
		for (long i = 2; i * i <= x; i++) {
			int cnt = 0;
			while (x % i == 0) {
				cnt++;
				x /= i;
			}
			if (cnt > 0) {
				factors.add(new long[] { i, cnt });
			}
		}
		if (x != 1) {
			factors.add(new long[] { x, 1 });
		}

	}

	public static long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int t = sc.nextInt();
		loop: while (t-- > 0) {
			long p = sc.nextLong();
			long q = sc.nextLong();
			getFactors(q);
//			getDiv(0, 1);
			long x = 1;
			for (long[] factor : factors) {
				int cnt = 0;
				long temp = p;
				while (temp % factor[0] == 0) {
					temp /= factor[0];
					cnt++;
				}
				long need = Math.max(cnt - (factor[1] - 1), 0);
				long h = p;
				while (need-- > 0) {
					h /= factor[0];
				}
				x = Math.max(x, h);
			}
			pw.println(x);
		}
		pw.close();

	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(FileReader f) {
			br = new BufferedReader(f);
		}

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

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

		public int[] nextIntArr(int n) throws IOException {
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = Integer.parseInt(next());
			}
			return arr;
		}

	}

}
