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
import sys
input = sys.stdin.readline
def inlist():
    return(list(map(int,input().split())))
def Sieve():
    isp = []
    mxN = 100000
    for i in range(0 , mxN):
        isp.append(0)
    for i in range(4 , mxN , 2):
        isp[i] = 1
    sq = int(math.sqrt(mxN))
    for i in range(3 , sq + 1, 2):
        if isp[i] == 0:
            for j in range(i * i , mxN , i + i):
                isp[j] = 1
    prime = []
    prime.append(2)
    for i in range(3 , mxN , 2):
        if isp[i] == 0:
            prime.append(i)
    return prime
def main():
    prime = Sieve()
    t = int(input())
    for tc in range(t):
        pq = inlist()
        p = pq[0]
        q = pq[1]
        x = 1
        if p < q:
            x = p
        else:
            if p % q != 0:
                x = p
            else:
                x = 1
                for i in prime:
                    if q == 1:
                        break
                    if q % i == 0:
                        tmp = p
                        cnt = 0
                        while q % i == 0:
                            q //= i
                            cnt += 1
                        tot = 0
                        while tmp % i == 0:
                            tmp //= i
                        cnt -= 1
                        while cnt != 0:
                            tmp *= i
                            cnt -= 1
                        x = max(x , tmp)
                if q != 1:
                    while p % q == 0:
                        p //= q
                    x = max(x , p)
        sys.stdout.write(str(x) + "\n")
main()