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.
"""1406 Problem B
"""
import math
from sys import stdin,stdout

def getInt():
	return int(stdin.readline())

def getInts():
	return list(map(int,stdin.readline().split()))

def solve():
  n = getInt()
  ab = getInts()
  minus = []
  add = []
  x = 0
  for i in range(n):
    if ab[i] < 0:
      minus.append(ab[i])
    elif ab[i] > 0:
      add.append(ab[i])
    else:
      x = 1
  minus.sort()
  add.sort()

  has1 = [0]*6
  has2 = [0]*6
     

  has1[0] = 1
  has2[0] = 1

  for i in range(1,6):
    if i > len(minus):
      break
    if i%2 == 1:
        minus.reverse()
        mul = 1
        for j in range(i):
                mul = mul * minus[j]
        has1[i] = mul
        minus.reverse()
    else:
            mul = 1
            for j in range(i):
                mul = mul * minus[j]
            has1[i] = mul
    
  mul = 1
  add.reverse()

  for i in range(len(add)):
    if i+1 > 5:
      break
    mul = mul * add[i]
    has2[i+1] = mul 
    
  ans = 1
    
  if x > 0:
        ans = 0
  else:
        ans = int(-1e18)

  for i in range(6):
            if has1[i] == 0 or has2[5-i] == 0 :
                continue
            else:
                ans = max(ans, has1[i] * has2[5-i])

  return ans

def main():
	t = getInt()
	for i in range(t):
		stdout.write(str(solve()) + "\n")


if __name__ == "__main__":
	main()