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.
t = int(input())
def getSmallestPrimeFac(n):
    s = int(pow(n,0.5))
    n2 = n
    count=0
    for i in range(2,s+1):
        
        if n%i==0:
            while n%i==0:
                count+=1 
                n = n//i
                
            return i,count
        return n,1


def reduce(n,d):
    s = 0
    while n%d==0:
        #print(n,d)
        n //= d
        s += 1
    return n,s

def solve(n,m):
    s = int(m**0.5)
    orig_n = n
    orig_m = m
    
    d_min = orig_n
    for i in range(2,s+1):
       
        

        count_n = 0
        count_m = 0
        n,count_n = reduce(n,i)
        m,count_m = reduce(m,i)


        if count_m>count_n:
            return 1
        
        #print('lala',i)
        if count_n>=count_m:
            
            a = i**(count_n - count_m + 1)
            if count_m>0:
                if a<d_min:
                    d_min = a

    
    if m>1:
        n,count_n = reduce(n,m)
        if count_n!=0:
            return min(d_min,m**(count_n))
        return 1

    return d_min



for _ in range(0,t):

    p,q = [int(x) for x in input().split(' ')]
        
    d_min=solve(p,q)
    print(p//d_min)





