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.
#!/usr/bin/env python
# coding: utf-8

# In[5]:


'''
T=int(input())
def maxProductSubarrayOfSizeK(A, n, k): 

    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 

    i = 0

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

    for itr in range( k) : 

        left_product = A[i] * A[i + 1] 

        right_product = A[j] * A[j - 1] 

        if (left_product > right_product) : 
            product *= left_product 
            i += 2
          
        else : 
            product *= right_product 
            j -= 2

    return product

for i in range(T):
    N=int(input())
    x=list(map(int,input().split()))
    if N==5:
        print(x[0]*x[1]*x[2]*x[4]*x[3])
    else:
        ans= maxProductSubarrayOfSizeK(x, N, 5)
        print(ans)
        
        
'''     


# In[1]:


ko=10**18
t=int(input())
for you in range(t):
    n=int(input())
    l=input().split()
    li=[int(i) for i in l]
    neg=[]
    pos=[]
    for i in li:
        if(i<0):
            neg.append(i)
        else:
            pos.append(i)
    neg.sort()
    pos.sort()
    z=min(5,len(neg))
    maxa=-ko
    yo=len(pos)
    for i in range(z+1):
        ans=1
        for j in range(i):
            ans=ans*neg[j]
        if(yo<5-i):
            continue
        for k in range(5-i):
            ans=ans*pos[-(k+1)]
        maxa=max(maxa,ans)
    for i in range(z+1):
        ans=1
        for j in range(i):
            ans=ans*neg[-(j+1)]
        if(yo<5-i):
            continue
        for k in range(5-i):
            ans=ans*pos[k]
        maxa=max(maxa,ans)
    print(maxa)


# In[ ]:




