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.
from sys import stdin,stdout
stdin.readline
def mp(): return list(map(int, stdin.readline().strip().split()))
def it():return int(stdin.readline().strip())
from collections import defaultdict as dd,Counter as C

def pf(n):
	v=[]
	d = dd(lambda:0)
	while n%2==0:
		d[2] +=1
		n//=2
	i=3
	while i*i<=n:
		if n%i==0:
			d[i]+=1
			n//=i
		else:
			i+=1
	if n>2:
		d[n]+=1
	return dict(d)

for _ in range(it()):
	p,q = mp()
	d = pf(q)
	ans = q
	ans =1
	for i,j in d.items():
		k = p
		while k%q == 0:
			k//=i
		ans = max(ans,k)
	print(ans)
