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.
gans = []
for _ in range(int(input())):
    p, q = map(int, input().split())
    q0 = q
    if p % q != 0:
        gans.append(p)
        continue
    i = 2
    u = []
    while i * i <= q:
        if q % i == 0:
            u.append(i)
        while q % i == 0:
            q //= i
        i += 1
    if q != 1:
        u.append(q)
    ans = p
    q = q0
    for i in range(len(u)):
        cur = 1
        while (p % cur == 0) and ((p // cur) % q == 0):
            cur *= u[i]
            #print(cur, p, q, (p // cur) % q, p // cur)
        if p % cur == 0:
            ans = min(ans, cur)
    gans.append(p // ans)
print(*gans, sep='\n')
