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 smallestDivisor(n): 
  
    if (n % 2 == 0): 
        return 2; 
    i = 3;  
    while(i * i <= n): 
        if (n % i == 0): 
            return i; 
        i += 2; 
  
    return n; 
  
def prevPowerofK(n, k):
 
    p = int(math.log(n) / math.log(k))
    return int(math.pow(k, p))

def primeFactors(n): 
    l = []
    while n % 2 == 0: 
        l.append(2) 
        n = n // 2
          
    for i in range(3,int(math.sqrt(n))+1,2): 
        while n % i== 0: 
            l.append(i)
            n = n // i 
             
    if n > 2: 
        l.append(n)
    return l
t = int(input())
while t:
    n,x= map(int,input().split())
    if n%x!=0:
        print(n)
    else:
        l2 = primeFactors(x)
        
        
        d2 = {}
        for j in l2:
            if j in d2:
                d2[j]+=1
            else:
                d2[j]=1
        
        ans = 0
        for i in d2:
            if n%i==0:
                power = 1
                number = i
                while n%number==0:
                    power+=1
                    number *=i
                if power>1:    
                    xxx = n// i**(power-1)
                    xxx *= i**(d2[i]-1)
                    ans = max(ans,xxx)
        print(ans)

    
    t-=1
    
    