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.
def maxmulSubarrayOfSizeK(A, n, k): 
	
    mul = 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): 
            mul *= A[i] 
        return mul 

    i = 0
  
    j = n - 1
  
 
    if (k & 1): 
        mul *= A[j] 
        j-= 1
        k-=1
  
    k >>= 1

    for itr in range( k) : 
  
        # mul from left pointers 
        left_mul = A[i] * A[i + 1] 
  
        # mul from right pointers 
        right_mul = A[j] * A[j - 1] 

        if (left_mul > right_mul) : 
            mul *= left_mul 
            i += 2
          
        else : 
            mul *= right_mul 
            j -= 2
  
    return mul 

for _ in range(int(input())):
    
    n = int(input())
    
    a = list(map(int,input().split()))
    ng = 0
    for i in a:
        if i < 0:
            ng += 1
    a.sort()
    if ng == n:
        ans = 1
        for i in range(n-1,n-6,-1):
            ans *= a[i]
        print(ans)
    else:
        print(maxmulSubarrayOfSizeK(a,n,5))
    
    