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.
# Author : Pratyaydeep↯Ghanta
import sys
inp=sys.stdin.buffer.readline
inar=lambda: list(map(int,inp().split()))
inin=lambda: int(inp())
inst=lambda: inp().decode().rstrip('\n\r')
# Am I debugging, check if I'm using same variable name in two places
from functools import reduce
def mul(a):
    if a==[]:
        return 1
    return reduce(lambda x,y:x*y, a)
_T_=inin()
for _t_ in range(_T_):
    n=inin()
    ar=inar()
    pos=[]
    neg=[]
    for i in ar:
        if i<0:
            neg.append(i)
        else:
            pos.append(i)
    neg.sort()
    pos.sort()
    #print(neg)
    #print(pos)
    #print(mul(neg))
    ans=-float('inf')
    for k in range(6):
        p=k
        ne=5-k
        if len(pos)>=p and len(neg)>=ne:
            if ne%2==0:
               ans=max(ans,mul(pos[-p:])*mul(neg[:ne]))
            else:
                ans=max(ans,mul(pos[-p:])*mul(neg[-ne:]))
                #print(neg[:ne])
    print(ans)