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.
t = int(input())
for i in range(0,t):
    l1 = [int(x) for x in input().split()]
    p = l1[0]
    q = l1[1]
    #all the factors of q:
    v1 = []
    v2 = []
    for j in range(1,int(q**0.5)+1):
        if q%j==0:
            if q//j==j:
                v1.append(j)
            else:
                v1.append(j)
                v2.append(q//j)
    for j in range(len(v2)-1,-1,-1):
        v1.append(v2[j])
    if p%q!=0:
        print(p)
    else:
        j = 0
        max_val = 0
        while True and j<len(v1):
            out = False
            if v1[j]==1:
                j+=1
                continue
            p_dash = p
            while p_dash%v1[j]==0:
                p_dash//=v1[j]
                if p_dash%q!=0:
                    if p_dash>max_val:
                        max_val = p_dash
                    out = True
            j+=1
        print(max_val)