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.
from sys import stdin
from collections import *


def factorize(n):  # o(sqr(n))
    c, ans = 1, []
    while (c * c < n):
        if n % c == 0:
            ans.extend([c, n // c])
        c += 1

    if c * c == n:
        ans.append(c)
    return sorted(ans)[1:]


for _ in range(int(input())):
    p, q = map(int, stdin.readline().split())
    if p % q:
        print(p)
    else:
        fac, ans, q1, p1, mem, mem1 = factorize(q), 0, q, p, defaultdict(int), defaultdict(int)

        for i in fac:
            while q1 % i == 0:
                mem[i] += 1
                q1 //= i

            while p1 % i == 0:
                mem1[i] += 1
                p1 //= i

        for i, j in mem.items():
            ans = max(ans, (p // pow(i, mem1[i])) * pow(i, j - 1))

        print(ans)
