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

def primeFactors(n,s):
    while n % 2 == 0:
        try:
            s[2]+=1
        except:
            s[2]=1
        n = n // 2

    for i in range(3, int(math.sqrt(n))+1, 2):

        while n % i == 0:
            try:
                s[i]+=1
            except:
                s[i]=1
            n = n // i

    if n > 2:
        s[n]=1
    return s


# n = 315
# primeFactors(n)

for _ in range(int(input())):
    s={}
    a,b=map(int,input().split())
    if a%b!=0:
        print(a)
    else:
        s=primeFactors(b,s)
        ans=[]
        for i in s:
            temp=a
            cnt=0
            while temp%i==0 and temp>0:
                cnt+=1
                temp//=i
            # print(i,s[i],cnt)
            ans.append(i**(cnt-s[i]+1))
        # print(s,ans)
        print(int(a//min(ans)))
            


