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())
for tt in range(t):
  n = int(input())
  numbers = [int(x) for x in input().split()]
  N = []
  P = []
  Z = False
  # Build N, P and Z
  for x in numbers:
    if x > 0:
      P.append(x)
    elif x < 0:
      N.append(x)
    else:
      # x must be 0.
      Z = True
  # Sort the arrays, this takes O(n log n) time.
  P.sort()
  N.sort()
  # -10**18 is chosen because it's smaller than the smallest possible
  # answer of the problem (since the numbers are between -3000 and +3000)
  ans = -10**18
  # Try all the cases
  if Z:
    ans = max(ans, 0)
  if len(P) >= 5:
    ans = max(ans, P[-5]*P[-4]*P[-3]*P[-2]*P[-1])
  if len(P) >= 4 and len(N) >= 1:
    ans = max(ans, N[-1]*P[0]*P[1]*P[2]*P[3])
  if len(P) >= 3 and len(N) >= 2:
    ans = max(ans, P[-3]*P[-2]*P[-1]*N[0]*N[1])
  if len(P) >= 2 and len(N) >= 3:
    ans = max(ans, N[-3]*N[-2]*N[-1]*P[0]*P[1])
  if len(P) >= 1 and len(N) >= 4:
    ans = max(ans, P[-1]*N[0]*N[1]*N[2]*N[3])
  if len(N) >= 5:
    ans = max(ans, N[-5]*N[-4]*N[-3]*N[-2]*N[-1])
  print(ans)
