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.
# Python program to print prime factors

import math

# A function to print all prime factors of
# a given number n
def primeFactors(n,d):
	
	# Print the number of two's that divide n
	if(n%2==0):
	    d[2]=0
	while n % 2 == 0:
		d[2]+=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):
		    d[i]=0
		while n % i== 0:
			d[i]+=1
			n = n // i
			
	# Condition if n is a prime
	# number greater than 2
	if n > 2:
		d[n]=1


t = int(input())
for i in range(t):
    p,q = map(int,input().split())
    if(p%q!=0):
        print(p)
    else:
        d = {}
        d1 = {}
        primeFactors(q,d)
        for key in d:
            p1 = p+0
            if(p1%key==0):
                d1[key]=0
                while(p1%key==0):
                    d1[key]+=1
                    p1//=key
        l = []
        for key in d:
            l.append(pow(key,d1[key]-(d[key]-1)))
        print(p//min(l))