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.
import sys

def prod(arr,num,p):

	l=len(arr)

	pro=1

	if(p==0):
		for h in range(num):
			pro*=arr[h]
		return pro
	else:
		ctr=0
		for h in range(l-1,-1,-1):
			if(ctr<num):
				pro*=arr[h]
			ctr+=1
		return pro


def solve(n,a):

	pos=[]
	neg=[]
	zer=[]

	for h in range(n):
		if(a[h]>0):
			pos.append(a[h])
		elif(a[h]<0):
			neg.append(a[h])
		else:
			zer.append(a[h])

	pos.sort()
	neg.sort()	
		
	ans=0
	flag=False

	pl = len(pos)
	nl = len(neg)
	zl = len(zer)

	# CASE 1
	if(pl>=5):
		val = prod(pos,5,1)
		
		if(ans==False):
			ans=val
			flag=True
		else:
			ans=max(ans,val)

	if(nl>=2 and pl>=3):
		val = prod(pos,3,1)*prod(neg,2,0)

		if(ans==False):
			ans=val
			flag=True
		else:
			ans=max(ans,val)

	if(nl>=4 and pl>=1):
		val = prod(pos,1,1)*prod(neg,4,0)

		if(ans==False):
			ans=val
			flag=True
		else:
			ans=max(ans,val)

	# CASE 2
	if(zl>0):
		ans = max(ans,0)

	# CASE 3
	ab_array=[]

	for e in range(n):
		ab_array.append(abs(a[e]))

	ab_array.sort()
	
	val = (-1)*prod(ab_array,5,0)

	if(ans==False):
		ans=val
		flag=True
	else:
		ans=max(ans,val)

	return ans

t=int(sys.stdin.readline())

ans_arr=[]

for i in range(t):
	n=int(sys.stdin.readline())

	a=[int(j) for j in sys.stdin.readline().split()]

	answer = solve(n,a)

	ans_arr.append(str(answer))

print("\n".join(ans_arr))