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.
from collections import defaultdict as dd
import sys
input=sys.stdin.readline
t=int(input())
while t:
    n=int(input())
    #n,m=map(int,input().split())
    l=list(map(int,input().split()))
    ll=[]
    for i in range(n):
        if(l[i]<0):
            ll.append((-l[i],1,l[i]))
        else:
            ll.append((l[i],0,l[i]))
    ll.sort(reverse=True)
    mul=1
    pm=10000000000000
    pmm=pm
    nm=-1000000000000
    for i in range(5):
        if(ll[i][2]>0):
            pm=min(pm,ll[i][2])
        if(ll[i][2]<0):
            nm=max(nm,ll[i][2])
        mul*=ll[i][2]
    mmul=mul
    lol=0
    ans=mul
    for i in range(5,n):
        if(ll[i][2]>=0 and mul<0):
            mmul=mul//nm
            mmul=mmul*ll[i][2]
            ans=max(ans,mmul)
        if(ll[i][2]<0 and pm!=pmm):
            mmul=mul//pm
            mmul=mmul*ll[i][2]
            ans=max(ans,mmul)
    if(ans<0):
        cou=0
        ans=1
        for i in range(n-1,-1,-1):
            ans=ans*ll[i][2]
            cou+=1
            if(cou==5):
                break
        
    print(ans)
    t-=1
