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.
# ------------
for _ in range(int(input())):
    a,b = list(map(int,input().split()))
    if a%b != 0:
        print(a)
        continue
    bDevVal = []
    bDevAm = []
    iter = 2
    bc = b
    while iter*iter <= bc:
        if bc%iter == 0:
            bDevVal.append(iter)
            bDevAm.append(0)
        while bc%iter == 0:
            bDevAm[-1]+=1
            bc//=iter
        iter+=1
    # print(bDevVal)
    # print(bDevAm)
    if bc > 1:
        bDevVal.append(bc)
        bDevAm.append(1)
    m = 1
    for i in range(len(bDevVal)):
        ac = a
        while ac%bDevVal[i] == 0:
            ac//=bDevVal[i]
        ac*=(bDevVal[i]**(bDevAm[i]-1))
        m = max(ac,m)
    print(m)