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 sys import stdin,stdout
from math import gcd
nmbr=lambda:int(stdin.readline())
lst=lambda:list(map(int, stdin.readline().split()))
def fact(n):
    mp={}
    while n%2==0:
        mp[2]=mp.get(2,0)+1
        n//=2
    div=3
    while div*div<=n:
        while n%div==0:
            mp[div]=mp.get(div,0)+1
            n//=div
        div+=2
    if n>2:mp[n]=1
    return mp
for _ in range(1):#nmbr()):
    tc=nmbr()
    for _ in range(tc):
        p,q=lst()
        if p%q!=0:print(p)
        else:
            ans=float('inf')
            # fp=fact(p)
            fq=fact(q)
            for f,pow in fq.items():
                n=p
                power=0
                while n%f==0:
                    n//=f
                    power+=1
                extra=power-(pow-1)
                ans=min(ans,f**extra)
            print(p//ans)