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
import sys,bisect
from heapq import *
from itertools import *
from collections import *
sys.setrecursionlimit(10 ** 6)
I = lambda : int(sys.stdin.readline())
neo = lambda : map(int, sys.stdin.readline().split())
Neo = lambda : list(map(int, sys.stdin.readline().split()))
def factorize(n): 
    count = 0
    d = defaultdict(int)
    while ((n % 2 > 0) == False):  
        n >>= 1
        count += 1
    if count > 0:    
        d[2] += count
    for i in range(3, int(math.sqrt(n)) + 1): 
        count = 0
        while (n % i == 0):  
            count += 1; 
            n = n//i
        if count > 0:    
            d[i] += count 
        i += 2
    if (n > 2): 
        d[n] += 1
    return d    
for _ in range(I()):
    p, q = neo()
    d,Ans = {},[]
    if p % q != 0:
        print(p)
        continue    
    d = factorize(q)
    Ans = 0
    for i,j in d.items():
        t = p
        while t%i == 0:
            t = t//i  
        Ans = max(Ans,t*i**(j-1))
    print(Ans)       