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


# Function to calculate all the prime
# factors and count of every prime factor
def factorize(n):
    count = 0
    res=[]

    # count the number of
    # times 2 divides
    while ((n % 2 > 0) == False):
        # equivalent to n = n / 2;
        n >>= 1
        count += 1

        # if 2 divides it
    if (count > 0):
        res.append([2,count])

        # check for all the possible
    # numbers that can divide it
    for i in range(3, int(math.sqrt(n)) + 1):
        count = 0
        while (n % i == 0):
            count += 1
            n = int(n / i)
        if (count > 0):
            res.append([i,count])
        i += 2

        # if n at the end is a prime number.
    if (n > 2):
        res.append([n,1])

    # Driver Code
    return res



















t=int(input())
for _ in range(t):
    p,q=map(int,input().split())
    if q>p:
        print(p)

    else:
        if p%q!=0:
            print(p)

        else:
            r=p//q
            req1=factorize(q)
            req1.sort()
            res=1

            for i in range(len(req1)):
                cnt=0

                j=req1[i][0]
                while(r%(j**(cnt+1))==0):
                    cnt+=1
                if cnt:
                    if req1[i][1]>cnt:
                        res=max(res,r*(q//(req1[i][0]**req1[i][1])))
                        res=max(res,p//(req1[i][0]**(cnt+1)))

                    else:
                        res=max(res,(r*q)//(req1[i][0]**(cnt+1)))
                        if len(req1)>1:
                            res=max(res,req1[i][0]**(cnt+req1[i][1]))



                else:
                    res=max(res,(q//req1[i][0])*(r))

            print(res)






