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.
# NTFS: galen_colin
## Suppose q = 2^x*3^y*5^z then we can remove all 2's and multiply (x-1) 2's in p
## so that it is not divisible by q.
## Do the same for y,z too and take max of them
from collections import defaultdict


# prime factorization seive
def seive(n):
    dic = defaultdict(int)
    i = 2
    tmp = n
    while i * i <= tmp:
        count = 0
        while n % i == 0:
            n = n // i
            count += 1
        if count:
            dic[i] = count
        i += 1
    if n - 1:
        dic[n] = 1
    return dic


for _ in range(int(input())):
    p, q = map(int, input().split())
    if p % q:
        print(p)
        continue
    factors = seive(q)
    ans = 1
    for fact in factors:  # Iterating over all factors
        tmp = p  # Starting with p remove all facts
        while tmp % fact == 0:
            tmp = tmp // fact
        tmp = tmp * (fact**(factors[fact] - 1))  # multiply (x-1) facts again
        ans = max(ans, tmp)  # take max of all facts
    print(ans)