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.
t=int(input())
for i in range(t):
    n=int(input())
    lis=list(map(int,input().split()))
    lis.sort()
    ng=0
    pg=0
    c=1
    
    for num in lis:
        if num<0:
            ng=ng+1
        else:
            pg=pg+1
    mult=1
    if pg==0:
        print(lis[n-1]*lis[n-2]*lis[n-3]*lis[n-4]*lis[n-5])
    elif pg==1:
        for j in range(4):
            mult=mult*lis[j]
        mult=mult*lis[n-1]
        print(mult)
    elif pg==2 and ng==3:
        for j in range(5):
            mult=mult*lis[j]
        print(mult)
    elif pg==2 and ng>=4:
        k1=1
        for j in range(4):
            k1=k1*lis[j]
        k1=k1*lis[n-1]        
        print(k1)
    elif pg==3 and ng<4:
        k1=1
        for j in range(2):
            k1=k1*lis[j]
        k1=k1*lis[n-1]*lis[n-2]*lis[n-3]
        print(k1)
    elif pg==3 and ng>=4:
        k1=1
        k2=1
        for j in range(4):
            k1=k1*lis[j]
        k1=k1*lis[n-1]   
        for j in range(2):
            k2=k2*lis[j]
        k2=k2*lis[n-1]*lis[n-2]*lis[n-3]
        print(max(k1,k2))
    elif pg==4 and ng==1:
        for j in range(5):
            mult=mult*lis[j]
        print(mult)
    elif pg==4 and (ng==2 or ng==3):
        k2=1
        for j in range(2):
            k2=k2*lis[j]
        k2=k2*lis[n-1]*lis[n-2]*lis[n-3]
        print(k2)
    elif pg==4 and ng>=4:
        k1=1
        k2=1
        for j in range(4):
            k1=k1*lis[j]
        k1=k1*lis[n-1]   
        for j in range(2):
            k2=k2*lis[j]
        k2=k2*lis[n-1]*lis[n-2]*lis[n-3]
        print(max(k1,k2))
    elif pg>=5 and ng>=2:
        k1=1
        k2=1
        k3=lis[n-1]*lis[n-2]*lis[n-3]*lis[n-4]*lis[n-5]
        for j in range(4):
            k1=k1*lis[j]
        k1=k1*lis[n-1]   
        for j in range(2):
            k2=k2*lis[j]
        k2=k2*lis[n-1]*lis[n-2]*lis[n-3]
        print(max(k1,k2,k3))
    else:
        print(lis[n-1]*lis[n-2]*lis[n-3]*lis[n-4]*lis[n-5])    
        
        
        
        
            
        
        