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 read_generator():
    while True:
        tokens = input().split(' ')
        for t in tokens:
            yield t

reader = read_generator()

def readword():
    return next(reader)

def readint():
    return int(next(reader))

def readfloat():
    return float(next(reader))

def readline():
    return input()

import math

def solve(a, b):
    if a < b:
        return a
    if a > b and a % b != 0:
        return a

    d = []
    if a % b == 0:
        d.append(b)

    for i in range(2, math.ceil(math.sqrt(b) + 1)):
        if b % i == 0:
            d.append(i)
            if a % (b // i) == 0:
                d.append(b // i)

    res = []
    for v in d:
        if v != 1:
            t = a
            while t % v == 0:
                t //= v
                if t % b != 0:
                    res.append(t)
                    break

    res.append(1)
    return max(res)

def solve2(a, b):
    for i in range(a, 0, -1):
        if a % i == 0 and i % b != 0:
            return i

tests = readint()

for _ in range(tests):

    a, b = readint(), readint()
    print(solve(a, b))

