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 maxsize, stdout, stdin, stderr
# mod = int(1e9 + 7)
import re  # can use multiple splits
tup = lambda: map(int, stdin.readline().split())
I = lambda: int(stdin.readline())
lint = lambda: [int(x) for x in stdin.readline().split()]
S = lambda: stdin.readline().replace('\n', '').strip()
def grid(r, c): return [lint() for i in range(r)]
def debug(*args, c=6): print('\033[3{}m'.format(c), *args, '\033[0m', file=stderr)
stpr = lambda x  : stdout.write(f'{x}' + '\n')
star = lambda x : print(' '.join(map(str , x)))
from collections import *
from math import *
for _ in range(I()):
    a , b = tup()
    t = b
    if a%b!=0:
        print(a)
        continue
    d = defaultdict(int)
    while b%2==0:
        b//=2
        d[2]+=1
    for i in range(3, int(b**0.5) +  1 , 2):
        while b%i==0:
            b//=i
            d[i]+=1
        #if i==1:break
    if b > 2:
        d[b]+=1
    ans = 1
    #print(d)
    for i in d:
        p = a
        cnta,cntb = 0,0
        while p%i==0:
            p//=i
            cnta+=1
        #print(cnta , d[i] + 1 )
        ans = max(ans , a// i**(cnta - d[i] + 1))
    print(ans)



















