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 collections import Counter


def read_int():
    return int(input())


def read_ints():
    return map(int, input().split(' '))


primes = []
is_prime = [True for _ in range(100000)]
is_prime[0] = is_prime[1] = False
for i in range(2, 100000):
    if is_prime[i]:
        primes.append(i)
    for j in range(i * i, 100000, i):
        is_prime[j] = False


t = read_int()
for case_num in range(t):
    p, q = read_ints()
    if p % q != 0:
        print(p)
    else:
        cnt = Counter()
        x = q
        for prime in primes:
            if x % prime == 0:
                while x % prime == 0:
                    x //= prime
                    cnt[prime] += 1
        if x != 1:
            cnt[x] += 1
        lo = int(1e18)
        y = p
        for key in cnt:
            curr = 0
            while y % key == 0:
                y //= key
                curr += 1
            lo = min(lo, pow(key, curr - cnt[key] + 1))
        print(p // lo)
