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.
def primeFactors(n):
    import math
    factors = dict()
    while n % 2 == 0:
        if 2 not in factors:
            factors[2] = 0
        factors[2] += 1
        n = n // 2

    for i in range(3, int(math.sqrt(n)) + 1, 2):

        while n % i == 0:
            if i not in factors:
                factors[i] = 0
            factors[i] += 1
            n = n // i

    if n > 2:
        if n not in factors:
            factors[n] = 0
        factors[n] += 1

    return factors


def testcase():

    p, q = list(map(int, input().split()))

    if p < q or p % q != 0:
        print(p)
        return

    qf = primeFactors(q)
    pf = dict()
    pt = p
    for f in qf:
        pf[f] = 0
        while pt % f == 0 and f > 0:
            pf[f] += 1
            pt //= f

    div = 10 ** 18

    for f in qf:
        if qf[f] <= pf[f]:
            val = f ** (pf[f] - qf[f] + 1)
            div = min(div, val)

    ans = p // div
    print(ans)


    return


import sys, os
if os.path.exists('input.txt'):
    sys.stdin = open('input.txt', 'r')
sys.setrecursionlimit(10 ** 5)
t = int(input())
for _ in range(t):
    testcase()