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 primeFactors(n):
	primes = []
	# Print the number of two's that divide n
	fact = []
	#print('n je')
	if n%2==0:
		primes.append(2)
		fact.append(1)
		n = n // 2
	#print('primes je')
	#print(primes)
	while n % 2 == 0:
		fact[0] +=1
		n = n // 2
          
    # n must be odd at this point 
    # so a skip of 2 ( i = i + 2) can be used 
	for i in range(3,int(math.sqrt(n))+1,2): 
		if n%i==0:
			primes.append(i)
			fact.append(1)
			n = n // i  
        # while i divides n , print i ad divide n 
		while n % i== 0:
			fact[-1] +=1
			n = n // i 
              
    # Condition if n is a prime 
    # number greater than 2 
	if n > 2:
		primes.append(n)
		fact.append(1)		
	return [primes, fact]		
		
import sys
input = sys.stdin.readline
t = int(input())
#print(primeFactors(t))
for test in range(t):
	#n = int(input())
	[p, q] = list(map(int, input().split(" ")))
	
	#a = list(map(int, input().split(" ")))
	[primes, fact] = primeFactors(q)
	#print(primes)
	#print(fact)
	#print(primeFactors(p))
	deli = p%q
	#print(int(p))
	if deli == 0:
		k = []
		for i in range(len(primes)):
			prox = p
			#print(prox)
			fac = 0
			#print(prox % primes[i])
			while prox % primes[i]== 0:
				fac +=1
				prox = prox //primes[i]
				#print(fac)
				#print(int(prox))
			k.append(primes[i]**(fac-fact[i]+1))
			#print(k)
		#print(k)
		m = min(k)
		#print(p)
		res = p//m
		print(res)	
	else:
		#print('a')
		print(int(p))