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 mul(arr):
    res = 1
    for e in arr:
        res *= e
    return res


if __name__ == '__main__':
    t = int(input())
    for _ in range(t):
        n = int(input())
        a = [int(x) for x in input().split()]
        res = []
        positive = sorted([e for e in a if e >= 0], reverse=True)
        negative = sorted([e for e in a if e < 0], reverse=False)
        for e in positive:
            if e == 0:
                res.append(0)
                break
        if len(positive) >= 5:
            res.append(mul(positive[:5]))
        if len(negative) >= 5:
            res.append(mul(negative[len(negative) - 5:]))
        for i in range(1, 5):
            if len(negative) >= 5 - i and len(positive) >= i:
                if (5 - i) % 2 == 0:
                    res.append(mul(negative[:5 - i]) * mul(positive[:i]))
                else:
                    res.append(mul(negative[len(negative) - 5 + i:]) * mul(positive[len(positive) - i:]))
        print(max(res))