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())
ans = list()

for i in range(t):
    ln = int(input())
    arr = list(map(int, input().split()))

    a=False
    for k in arr:
        if k>0:
            a = True
            break

    if a:
        arr.sort(reverse=True, key=lambda x:abs(x))
        numbers = list()
        mn = 1
        last_p = 0
        last_n = 0
        for k in range(5):
            numbers.append(arr[k])
            mn*=arr[k]
            if arr[k]>0:
                last_p = k
            else:
                last_n = k
        if mn>=0:
            ans.append(mn)
        else:
            mp = 0
            mz = 0
            for k in range(5, len(arr)):
                if arr[k]>0:
                    mp = k
                    break
            for k in range(5, len(arr)):
                if arr[k]<0:
                    mz = k
                    break
            l = int(mn/arr[last_p])*arr[mz]
            r = int(mn/arr[last_n])*arr[mp]
            if (mz>0 and mp>0 and l>= r) or (mz>0 and l>= r):
                mn = l
            elif (mz>0 and mp>0 and r>=l) or (mp>0 and r>=l):
                mn = r
            elif mz>0:
                mn = l
            elif mp>0:
                mn = r
            elif 0 in arr:
                mn = 0
            ans.append(mn)
    else:
        arr.sort(reverse=True)
        mn = 1
        for l in range(5):
            mn*=arr[l]
        ans.append(mn)
for l in ans:
    print(l)