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.
# Python 3 code to find maximum possible 
# product of sub-sequence of size k from 
# given array of n integers 

# Required function 
def maxProductSubarrayOfSizeK(A, n, k): 

	# sorting given input array 
	A.sort() 

	product = 1

	if (A[n - 1] == 0 and (k & 1)): 
		return 0

	if (A[n - 1] <= 0 and (k & 1)) : 
		for i in range(n - 1, n - k + 1, -1): 
			product *= A[i] 
		return product 

	# else 
	# i is current left pointer index 
	i = 0

	# j is current right pointer index 
	j = n - 1

	if (k & 1): 
		product *= A[j] 
		j-= 1
		k-=1

	k >>= 1

	# Now finding k corresponding pairs to get 
	# maximum possible value of product 
	for itr in range( k) : 

		# product from left pointers 
		left_product = A[i] * A[i + 1] 

		# product from right pointers 
		right_product = A[j] * A[j - 1] 

		# Taking the max product from two 
		# choices. Correspondingly changing 
		# the pointer's position 
		if (left_product > right_product) : 
			product *= left_product 
			i += 2
		
		else : 
			product *= right_product 
			j -= 2

	# Finally return product 
	return product 

# Driver Code 
if __name__ == "__main__": 
	
	for _ in range(int(input())):
	    n = int(input())
	    A = list(map(int, input().split()))
	    p = 0
	    for i in A:
	        if i >= 0:
	            p += 1
	            break
	    if p == 0:
	        A.sort(reverse = True)
	        B = A[0:5]
	        prod = 1
	        for i in B:
	            prod *= i
	        print(prod)
	    else:
	        print(maxProductSubarrayOfSizeK(A, n, 5)) 

