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.
import sys


def get_ints(): return list(map(int, sys.stdin.readline().strip().split()))


def fast_print(input): sys.stdout.write(str(input)+"\n")


def multiply(numbers):
    total = 1
    for x in numbers:
        total *= x
    return total


t = int(input())
while t:
    t -= 1
    n = int(input())
    a = get_ints()
    pos, neg = [], []
    zero = False
    for x in a:
        if x > 0:
            pos.append(x)
        else:
            neg.append(x)
        if x == 0:
            zero = True
    pos.sort(reverse=True)
    neg.sort()
    ans = float("-inf")
    p = len(pos)
    n = len(neg)
    for i in range(0, 6):
        k = min(i, p)
        l = min(5 - k, n)
        if k+l == 5:
            temp = multiply(pos[:k])
            if l == 1 or l == 3 or l == 5:
                ans = max(ans, temp * multiply(neg[-l:]))
            else:
                ans = max(ans, temp * multiply(neg[:l]))
    ans = 0 if zero and ans < 0 else ans
    fast_print(ans)
