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 defaultdict
def factorize(n):
    d=defaultdict(int)
    count = 0 
    while ((n % 2 > 0) == False):  
        n >>= 1  
        count += 1
    if (count > 0): 
        d[2]+=count 
    for i in range(3, int(n**0.5) + 1): 
        count = 0
        while (n % i == 0):  
            count += 1
            n//=i
        if (count > 0): 
            d[i]+=count 
        i += 2  
    if (n > 2): 
        d[n]+=1
    return d
for i in range(int(input())):
    p,q=map(int,input().split())
    d2=factorize(q)
    ans=1
    for j in d2:
        c=j**d2[j]
        if p%c:
            ans=p
            break
        a=p
        while a%c==0:
            a//=j
        ans=max(ans,a)
    print(ans)
