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.
import math
for _ in range(int(input())):
    n=int(input())
    a=list(map(int,input().split()))
    an,ap=[],[]
    kn,kp,ko=0,0,0
    for i in range(n):
        if a[i]<0:
            an.append(a[i])
            kn+=1
        elif a[i]>0:
            ap.append(a[i])
            kp+=1
        else:
            ko+=1
    an.sort()
    ap.sort()
    ap.reverse()
    check=1
    ans1=(-1)*float('inf')
    ans2=(-1)*float('inf')
    ans3=(-1)*float('inf')
    ans4=(-1)*float('inf')
    ans5=(-1)*float('inf')
    ans6=(-1)*float('inf')
    if kp>=5:
        ans1=ap[0]*ap[1]*ap[2]*ap[3]*ap[4]
    if kp>=4 and kn>=1:
        ans2=ap[kp-1]*ap[kp-2]*ap[kp-3]*ap[kp-4]*an[kn-1]
    if kp>=3 and kn>=2:
        ans3=ap[0]*ap[1]*ap[2]*an[0]*an[1]
    if kp>=2 and kn>=3:
        ans4=ap[kp-1]*ap[kp-2]*an[kn-1]*an[kn-2]*an[kn-3]
    if kp>=1 and kn>=4:
        ans5=ap[0]*an[0]*an[1]*an[2]*an[3]
    if kn>=5:
        ans6=an[kn-1]*an[kn-2]*an[kn-3]*an[kn-4]*an[kn-5]
    ans=max(ans1,ans2,ans3,ans4,ans5,ans6)
    if ko>0:
        print(max(ans,0))
    else:
        print(ans)