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.
def getPrimeFactorCounts(x): #returns #[[prime factor,cnts],...]
    primeFactorCnts=[] #[prime,cnts]
    for i in range(2,int(x**0.5)+1):
        cnt=0
        while x%i==0:
            cnt+=1
            x//=i
        if cnt>0:
            primeFactorCnts.append((i,cnt))
    if x>1:
        primeFactorCnts.append((x,1))
    return primeFactorCnts


t=int(input())
for _ in range(t):
    p,q=[int(x) for x in input().split()]
    
    qf=getPrimeFactorCounts(q)
    
    qpf=[] #primefactors of q
    for f,c in qf:
        qpf.append(f)
    
    ans=-1
    for f in qpf:
        p2=p
        while True:
            if p2%q!=0:
                ans=max(ans,p2)
                break
            if p2%f==0:
                p2//=f
            else:
                break
        
    print(ans)