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 math import floor,sqrt
t=int(input())
primes=[]
powers=[]
def factorization(n):
    for i in range(2,floor(sqrt(n)+1)):
        if n%i==0:
            power=0
            while n%i==0:
                power=power+1
                n=n//i
            primes.append(i)
            powers.append(power)
    if n!=1:
        primes.append(n)
        powers.append(1)
def vp(x,prime):
    powers2=[]
    for i in prime:
        c=1
        while x!=1 and x%i==0:
            x=x//i
            c=c+1
        powers2.append(i**c)
    return powers2
for i in range(t):
    p,q=input().split()
    p=int(p)
    q=int(q)
    if p<q:
        print(p)
    elif p%q!=0:
        print(p)
    else:
        primes=[]
        powers=[]
        factorization(q)
        s=p//q
        powers2=vp(s,primes)
        print(p//min(powers2))

