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,math,itertools
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right 
from heapq import heappop,heappush,heapify
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))

def prime_factorize(n):
    a = defaultdict(int)
    if n == 1:
        a[1] = 1
        return a
    while n % 2 == 0:
        a[2] += 1
        n //= 2
    f = 3
    while f * f <= n:
        if n % f == 0:
            a[f] += 1
            n //= f
        else:
            f += 2
    if n != 1:
        a[n] += 1
    return a
def a_count(key):
    n = p
    cnt = 0
    while n%key == 0:
        cnt += 1
        n //= key
    return cnt
t = inp()
for _ in range(t):
    p,q = inpl()
    if p%q: print(p); continue
    b = prime_factorize(q)
    res = 0
    for key in list(b):
        df = a_count(key)
        cnt = df-(b[key]-1)
        now = p//pow(key,cnt)
        res = max(res, now)
    print(res)
    
