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 sys
import math
from collections import defaultdict,Counter

input=sys.stdin.readline
def print(x):
    sys.stdout.write(str(x)+"\n")

# sys.stdout=open("CP1/output.txt",'w')
# sys.stdin=open("CP1/input.txt",'r')

# mod=pow(10,9)+7
def find(n):
	d=defaultdict(int)
	while n%2==0:
		n//=2
		d[2]+=1
	for j in range(3,int(math.sqrt(n))+1,2):
		while n%j==0:
			n//=j
			d[j]+=1
	if n>2:
		d[n]=1

	return d
t=int(input())
for i in range(t):
	p,q=map(int,input().split())
	if p%q!=0:
		print(p)
	else:
		p1=p
		d1=find(q)
		ans=10**18
		for j in d1:
			c=0
			while p%j==0:
				p//=j
				c+=1
			cur=pow(j,c-(d1[j]-1))
			ans=min(ans,cur)
		print(p1//ans)