Tomorrow is a difficult day for Polycarp: he has to attend a lectures and b practical classes at the university! Since Polycarp is a diligent student, he is going to attend all of them.

While preparing for the university, Polycarp wonders whether he can take enough writing implements to write all of the lectures and draw everything he has to during all of the practical classes. Polycarp writes lectures using a pen (he can't use a pencil to write lectures!); he can write down c lectures using one pen, and after that it runs out of ink. During practical classes Polycarp draws blueprints with a pencil (he can't use a pen to draw blueprints!); one pencil is enough to draw all blueprints during d practical classes, after which it is unusable.

Polycarp's pencilcase can hold no more than k writing implements, so if Polycarp wants to take x pens and y pencils, they will fit in the pencilcase if and only if x + y ≤ k.

Now Polycarp wants to know how many pens and pencils should he take. Help him to determine it, or tell that his pencilcase doesn't have enough room for all the implements he needs tomorrow!

Note that you don't have to minimize the number of writing implements (though their total number must not exceed k).

Input

The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then the test cases follow.

Each test case is described by one line containing five integers a, b, c, d and k, separated by spaces (1 ≤ a, b, c, d, k ≤ 100) — the number of lectures Polycarp has to attend, the number of practical classes Polycarp has to attend, the number of lectures which can be written down using one pen, the number of practical classes for which one pencil is enough, and the number of writing implements that can fit into Polycarp's pencilcase, respectively.

In hacks it is allowed to use only one test case in the input, so t = 1 should be satisfied.

Output

For each test case, print the answer as follows:

If the pencilcase can't hold enough writing implements to use them during all lectures and practical classes, print one integer -1. Otherwise, print two non-negative integers x and y — the number of pens and pencils Polycarp should put in his pencilcase. If there are multiple answers, print any of them. Note that you don't have to minimize the number of writing implements (though their total number must not exceed k).

Example

Input


3
7 5 4 5 8
7 5 4 5 2
20 53 45 26 4


Output


7 1
-1
1 3

Note

There are many different answers for the first test case; x = 7, y = 1 is only one of them. For example, x = 3, y = 1 is also correct.

x = 1, y = 3 is the only correct answer for the third test case.
''' CODED WITH LOVE BY SATYAM KUMAR '''

from sys import stdin, stdout
import heapq
import cProfile, math
from collections import Counter, defaultdict, deque
from bisect import bisect_left, bisect, bisect_right
import itertools
from copy import deepcopy
from fractions import Fraction
import sys, threading
import operator as op
from functools import reduce
import sys

sys.setrecursionlimit(10 ** 6)  # max depth of recursion
threading.stack_size(2 ** 27)  # new thread will get stack of such size
fac_warm_up = False
printHeap = str()
memory_constrained = False
P = 10 ** 9 + 7


class MergeFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.size = [1] * n
        self.num_sets = n
        self.lista = [[_] for _ in range(n)]

    def find(self, a):
        to_update = []
        while a != self.parent[a]:
            to_update.append(a)
            a = self.parent[a]
        for b in to_update:
            self.parent[b] = a
        return self.parent[a]

    def merge(self, a, b):
        a = self.find(a)
        b = self.find(b)
        if a == b:
            return
        if self.size[a] < self.size[b]:
            a, b = b, a
        self.num_sets -= 1
        self.parent[b] = a
        self.size[a] += self.size[b]
        self.lista[a] += self.lista[b]

    def set_size(self, a):
        return self.size[self.find(a)]

    def __len__(self):
        return self.num_sets


def display(string_to_print):
    stdout.write(str(string_to_print) + "\n")


def prime_factors(n):  # n**0.5 complex
    factors = dict()
    for i in range(2, math.ceil(math.sqrt(n)) + 1):
        while n % i == 0:
            if i in factors:
                factors[i] += 1
            else:
                factors[i] = 1
            n = n // i
    if n > 2:
        factors[n] = 1
    return (factors)


def all_factors(n):
    return set(reduce(list.__add__,
                      ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))


def fibonacci_modP(n, MOD):
    if n < 2: return 1
    return (cached_fn(fibonacci_modP, (n + 1) // 2, MOD) * cached_fn(fibonacci_modP, n // 2, MOD) + cached_fn(
        fibonacci_modP, (n - 1) // 2, MOD) * cached_fn(fibonacci_modP, (n - 2) // 2, MOD)) % MOD


def factorial_modP_Wilson(n, p):
    if (p <= n):
        return 0
    res = (p - 1)
    for i in range(n + 1, p):
        res = (res * cached_fn(InverseEuler, i, p)) % p
    return res


def binary(n, digits=20):
    b = bin(n)[2:]
    b = '0' * (digits - len(b)) + b
    return b


def is_prime(n):
    """Returns True if n is prime."""
    if n < 4:
        return True
    if n % 2 == 0:
        return False
    if n % 3 == 0:
        return False
    i = 5
    w = 2
    while i * i <= n:
        if n % i == 0:
            return False
        i += w
        w = 6 - w
    return True


def generate_primes(n):
    prime = [True for i in range(n + 1)]
    p = 2
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 1
    return prime


factorial_modP = []


def warm_up_fac(MOD):
    global factorial_modP, fac_warm_up
    if fac_warm_up: return
    factorial_modP = [1 for _ in range(fac_warm_up_size + 1)]
    for i in range(2, fac_warm_up_size):
        factorial_modP[i] = (factorial_modP[i - 1] * i) % MOD
    fac_warm_up = True


def InverseEuler(n, MOD):
    return pow(n, MOD - 2, MOD)


def nCr(n, r, MOD):
    global fac_warm_up, factorial_modP
    if not fac_warm_up:
        warm_up_fac(MOD)
        fac_warm_up = True
    return (factorial_modP[n] * (
            (pow(factorial_modP[r], MOD - 2, MOD) * pow(factorial_modP[n - r], MOD - 2, MOD)) % MOD)) % MOD


def get_int():
    return int(stdin.readline().strip())


def get_tuple():
    return map(int, stdin.readline().split())


def get_list():
    return list(map(int, stdin.readline().split()))


memory = dict()


def clear_cache():
    global memory
    memory = dict()


def cached_fn(fn, *args):
    global memory
    if args in memory:
        return memory[args]
    else:
        result = fn(*args)
        memory[args] = result
        return result


def ncr(n, r):
    return math.factorial(n) / (math.factorial(n - r) * math.factorial(r))


def binary_search(i, li):
    fn = lambda x: li[x] - x // i
    x = -1
    b = len(li)
    while b >= 1:
        while b + x < len(li) and fn(b + x) > 0:  # Change this condition 2 to whatever you like
            x += b
        b = b // 2
    return x


# -------------------------------------------------------------- MAIN PROGRAM


TestCases = True
optimise_for_recursion = True  # Can not be used clubbed with TestCases WHen using recursive functions, use Python 3


def main():
    a, b, c, d, k = get_tuple()
    x = (a+c-1)//c
    y = (b+d-1)//d
    if x+y<=k:
        print(x, y)
    else: print(-1)
# --------------------------------------------------------------------- END=


if TestCases:
    for i in range(get_int()):
        main()
else:
    main() if not optimise_for_recursion else threading.Thread(target=main).start()
