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.
# A. Division

import math
def prime_factors(a):
    s = int(math.sqrt(a))+1
    pfs = {}
    for i in range(2, s+1):
        if i > a:
            break
        ic = 0
        while a % i == 0:
            a //= i
            ic += 1
        if ic > 0:
            pfs[i] = ic
    if a > 1:
        pfs[a] = 1
    return pfs

t = int(input())
results = []
for i in range(t):
    pq = input().split()
    p = int(pq[0])
    q = int(pq[1])
    if p % q != 0:
        results.append(p)
    else:
        result = 1
        pfs = prime_factors(q)
        r = p // q
        pfs_2 = {}
        for k in pfs.keys():
            ic = 0
            while r % k == 0:
                r //= k
                ic += 1
            if ic > 0:
                pfs_2[k] = ic        
        for k, v1 in pfs.items():
            if k in pfs_2:
                v2 = pfs_2[k]
                prod = p // k ** (v2+1)
            else:
                prod = p // k
            if prod > result:
                result = prod
        results.append(result)


for result in results:
  print(result)