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.
def prod(x):
    res = 1
    for _ in x:
        res *= _
    return res

def solve(N, A):
    A = sorted(A)
    if A[-1] <= 0 or N == 5:
        print(prod(A[-5:]))
        return

    pos, neg = 0, 0
    for x in A:
        pos += (x > 0); neg += (x < 0)
    nol = N - (pos + neg)

    ans = -float('inf')
    if nol + pos >= 5:
        ans = max([ans, prod(A[-5:])])
    if nol + pos >= 3:
        ans = max([ans, prod(A[:2]) * prod(A[-3:])])
    if nol + pos >= 1:
        ans = max([ans, prod(A[:4]) * A[-1]])

    print(ans)

def main():
    T = int(input().strip())
    for _ in range(T):
        N = int(input().strip())
        A = list(map(int, input().strip().split()))
        solve(N, A)

if __name__ == '__main__':
    main()
