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 how_many_of_this_factor(p, factor):
    total = 0
    while p % factor == 0:
        p //= factor
        total += 1
    return total

def add_factor(factors, factor):
    if factor not in factors:
        factors[factor] = 1
    else:
        factors[factor] += 1

def factorize(n):
    factors = {}
    if n == 3:
        return {3 : 1}
    while n % 2 == 0:
        add_factor(factors, 2)
        n //= 2
    z = 3; hi = int(n ** 0.5)
    while z <= hi:
        got_one = False
        while n % z == 0:
            got_one = True
            n //= z
            add_factor(factors, z)
        if got_one:
            hi = int(n ** 0.5)
        z += 2
    if hi != 1 or n == 3:
        add_factor(factors, n)
    return factors

def solve(p, q):
    if p % q != 0:
        return p
    factors = factorize(q)
    m = p
    for factor in factors:
        p_has = how_many_of_this_factor(p, factor)
        q_has = factors[factor]
        toss = p_has - (q_has - 1)
        m = min(m, factor ** toss)
    return p // m

def main():
    t = int(input())
    for _ in range(t):
        p, q = map(int, input().split())
        print(solve(p, q))
main()