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.
from sys import stdin


def inp():
    return stdin.buffer.readline().rstrip().decode('utf8')


def itg():
    return int(stdin.buffer.readline())


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


# ############################## import
# ############################## main
from bisect import bisect_left
from functools import reduce
from operator import mul


def ml(array):
    return reduce(mul, array)

for __ in range(itg()):
    itg()
    arr = list(mpint())
    arr.sort()
    stack = [[], [], []]
    for a in arr:
        if a < 0:
            stack[0].append(a)
        elif a == 0:
            stack[1].append(0)
        else:
            stack[2].append(a)
    neg, zero, pos = stack
    neg.reverse()
    if len(neg) + len(pos) < 5:  # must contain 0
        print(0)
        continue
    if len(pos) == 0:  # ans must <= 0
        print(ml(neg[:5]) if not zero else 0)
        continue
    ans = -int(1e20) if not zero else 0
    if len(pos) >= 5:
        ans = ml(pos[-5:])
    if len(neg) >= 2 and len(pos) >= 3:  # --+++
        ans = max(ans, ml(neg[-2:]) * ml(pos[-3:]))
    if len(neg) >= 4 and pos:  # ----+
        ans = max(ans, ml(neg[-4:]) * pos[-1])
    if len(neg) == 1 and len(pos) >= 4:  # -++++
        ans = max(ans, neg[-1] * ml(pos[:4]))
    if len(neg) == 3 and len(pos) >= 2:  # ---++
        ans = max(ans, ml(neg[:3]) * ml(pos[:2]))
    print(ans)


# Please check!
