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 pf(n): 
    fac = {}
    while n % 2 == 0: 
        fac[2] = fac.get(2, 0)+1
        n = n // 2

    for i in range(3,int(math.sqrt(n))+1,2): 
        while n % i== 0: 
            fac[i] = fac.get(i, 0)+1
            n = n // i 

    if n > 2: 
        fac[n] = fac.get(n, 0)+1

    return fac

def gp(f, p):
    s = 1
    e = int(math.ceil(math.log(p,f)))

    val = 1
    while(s<=e):
        mid = (s+e)//2

        if p%(f**mid)!=0:
            e = mid - 1
        else:
            s = mid+1
            val = mid

    return val

test = int(input())
for _ in range(test):
    p, q = map(int, input().split())

    if p%q:
        print(p)
    else:
        pfac = pf(q)
        val = float("inf")
        for k in pfac:
            u = gp(k, p)
            val = min(val, (k**(u-pfac[k]+1)))
        ans = p//val
        print(ans)