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.
T = int(input().rstrip())

def get_max(posl, negl, npos, nneg):
    if len(posl) < npos or len(negl) < nneg:
        return None
    ans = 1
    if npos > 0:
        for i in posl[-npos:]:
            ans *= i
    if nneg > 0:
        for i in negl[-nneg:]:
            ans *= i
    return ans

def solve(arr):
    posl = []
    negl = []
    has_zero = False
    for i in arr:
        if i > 0:
            posl.append(i)
        elif i < 0:
            negl.append(i)
        else:
            has_zero = True
    posl.sort()
    negl.sort(reverse=True)
    all_max = [
        get_max(posl, negl, 5, 0),
        get_max(posl, negl, 3, 2),
        get_max(posl, negl, 1, 4)
    ]
    best_ans = all_max[0]
    for i in range(1, len(all_max)):
        curr = all_max[i]
        if curr is not None:
            if best_ans is None:
                best_ans = curr
            elif curr > best_ans:
                best_ans = curr

    if best_ans is not None:
        if best_ans > 0:
            return best_ans
    elif has_zero:
        return 0

    negl.reverse()
    all_min = [
        get_max(posl, negl, 4, 1),
        get_max(posl, negl, 2, 3),
        get_max(posl, negl, 0, 5)
    ]
    best_ans = all_min[0]
    for i in range(1, len(all_min)):
        curr = all_min[i]
        if curr is not None:
            if best_ans is None:
                best_ans = curr
            elif curr < best_ans:
                best_ans = curr
    return best_ans

for test in range(T):
    n = int(input().rstrip())
    arr = [int(x) for x in input().rstrip().split()]
    ret = solve(arr)
    print(ret)
