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.
import sys
if sys.subversion[0] == "PyPy":
    import io, atexit
    sys.stdout = io.BytesIO()
    atexit.register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
    
    sys.stdin = io.BytesIO(sys.stdin.read())
    input = lambda: sys.stdin.readline().rstrip()

RS = raw_input
RI = lambda x=int: map(x,RS().split())
RN = lambda x=int: x(RS())
''' ...................................................................... '''

def factorize(n):
    cnt = 0
    res = []
    while n&1==0:
        cnt+=1
        n/=2

    if cnt: res.append((2,cnt))

    lim = int(n**0.5)+2
    for i in xrange(3,lim,2):
        cnt = 0
        while n%i==0:
            cnt+=1
            n/=i
        if cnt: res.append((i,cnt))

    if n>1: res.append((n,1))
    return res

def solve(P,Q):
    if P%Q: return P

    ans = 0
    
    for (p,cnt) in factorize(Q):
        num = p**cnt

        tmp = P
        while tmp%num==0:
            tmp/=p

        ans = max(ans,tmp)
    return ans

for _ in xrange(RN()):
    P,Q = RI()
    print solve(P,Q)
