You are given an array of integers a_1,a_2,…,a_n. Find the maximum possible value of a_ia_ja_ka_la_t among all five indices (i, j, k, l, t) (i<j<k<l<t).

Input

The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n (5≤ n≤ 10^5) — the size of the array.

The second line of each test case contains n integers a_1,a_2,…,a_n (-3× 10^3≤ a_i≤ 3× 10^3) — given array.

It's guaranteed that the sum of n over all test cases does not exceed 2⋅ 10^5.

Output

For each test case, print one integer — the answer to the problem.

Example

Input


4
5
-1 -2 -3 -4 -5
6
-1 -2 -3 1 2 -1
6
-1 0 0 0 -1 -1
6
-9 -7 -5 -3 -2 1


Output


-120
12
0
945

Note

In the first test case, choosing a_1,a_2,a_3,a_4,a_5 is a best choice: (-1)⋅ (-2) ⋅ (-3)⋅ (-4)⋅ (-5)=-120.

In the second test case, choosing a_1,a_2,a_3,a_5,a_6 is a best choice: (-1)⋅ (-2) ⋅ (-3)⋅ 2⋅ (-1)=12.

In the third test case, choosing a_1,a_2,a_3,a_4,a_5 is a best choice: (-1)⋅ 0⋅ 0⋅ 0⋅ (-1)=0.

In the fourth test case, choosing a_1,a_2,a_3,a_4,a_6 is a best choice: (-9)⋅ (-7) ⋅ (-5)⋅ (-3)⋅ 1=945.
MOD = 1000000007
from collections import defaultdict as dd, Counter, deque


def si(): return input()


def ii(): return int(input())


def li(): return list(map(int, input().split()))


def mi(): return map(int, input().split())


def out(v): print(v)


def spout(): print(v, end=" ")


def d2b(n): return bin(n).replace("0b", "")


def twod(n, m, num): return [[num for x in range(m)] for y in range(n)]


def vow(): return ['a', 'e', 'i', 'o', 'u']


def let(): return [chr(i) for i in range(97, 123)]


def gcd(x, y):
    while y:
        x, y = y, x % y
    return x


def ispow2(x):
    return (x and (not (x & (x - 1))))


def prime_factors(n):
    i = 2
    factors = []
    while i * i <= n:
        if n % i:
            i += 1
        else:
            n //= i
            factors.append(i)
    if n > 1: factors.append(n)
    return (list(factors))
def prod(a):
    p=1
    for i in a: p*=i
    return p
t = ii()
while t:
    t -= 1
    n=ii()
    a=li()
    a.sort(reverse=True)
    ch=0
    if a[0]<0: ch=0
    elif a[-1]>0: ch=n-1
    else:
        while a[ch]>0: ch+=1
    if ch==0 or ch==n-1 or n==5: print(prod(a[0:5]))
    else:
        print(max(prod(a[0:5]),prod(a[0:3]+a[n-2:]),prod([a[0]]+a[n-4:])))