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 math
def primeFactors(n):
    d = {}
    while n % 2 == 0:
        d[2] = d.get(2,0) + 1
        n = n // 2

    for i in range(3, int(math.sqrt(n)) + 1, 2):
        while n % i == 0:
            d[i] = d.get(i,0) + 1
            n = n//i
    if n>2:
        d[n] = 1
    return d

def b_c(number, n):
    if n == 0:
        return 1
    else:
        if n%2 == 0:
            return b_c(number, n//2)*b_c(number, n//2)
        else:
            return number*b_c(number, (n-1)//2)*b_c(number, (n-1)//2)

def main():
    t = int(input())
    for __ in range(t):
        p, q = map(int, input().split())
        if p%q != 0:
            print(p)
        else:
            d = primeFactors(q)
            maxm = None
            for i in d.keys():
                x = p
                while x % q == 0:
                    x = x // i
                if maxm is None:
                    maxm = x
                else:
                    maxm = max(maxm, x)
            print(maxm)


if __name__ == '__main__':
    main()













