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
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log2, ceil
from collections import defaultdict as dd
from bisect import bisect_left as bl, bisect_right as br
from bisect import insort
from collections import Counter
from collections import deque
from heapq import heappush,heappop,heapify
from itertools import permutations,combinations
mod = int(1e9)+7
 
 
ip = lambda : int(stdin.readline())
inp = lambda: map(int,stdin.readline().split())
ips = lambda: stdin.readline().rstrip()
out = lambda x : stdout.write(str(x)+"\n")

def primeFactors(n):
    dic = dd(int)
    while n % 2 == 0: 
        dic[2] += 1 
        n = n // 2
    for i in range(3, int(math.sqrt(n))+1, 2): 
        while n % i == 0: 
            dic[i] += 1
            n = n // i 
    if n > 2: 
        dic[n] += 1
    return dic


t = ip()
for _ in range(t):
    p,q = inp()
    if p<q:
        ans = p
    else:
        if p%q != 0:
            ans = p
        else:
            g = gcd(p,q)
            if g == 1:
                ans = p
            else:
                prime = primeFactors(g)
                ch = sorted(list(prime.keys()))
                d = dd(int)
                for i in ch:
                    temp = p
                    while temp%i == 0:
                        d[i] += 1
                        temp = temp//i
                val = []
                for i in ch:
                    val.append(pow(i,d[i]-prime[i]+1))
                div = min(val)
                ans = p//div
    out(ans)
        
                    
    
    
    
    
    
    
            

    
            
            
            
        
    

            
        
