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 sys

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10**18
MOD = 10**9 + 7

for _ in range(INT()):
    N = INT()
    A = LIST()

    M = 5
    dp0 = list2d(N+1, M+1, -INF)
    dp1 = list2d(N+1, M+1, INF)
    dp0[0][0] = dp1[0][0] = 1
    for i in range(N):
        for j in range(M+1):
            if dp0[i][j] != -INF:
                dp0[i+1][j] = max(dp0[i+1][j], dp0[i][j])
                if j < M:
                    dp0[i+1][j+1] = max(dp0[i+1][j+1], dp0[i][j]*A[i])
                    dp1[i+1][j+1] = min(dp1[i+1][j+1], dp0[i][j]*A[i])
            if dp1[i][j] != INF:
                dp1[i+1][j] = min(dp1[i+1][j], dp1[i][j])
                if j < M:
                    dp1[i+1][j+1] = min(dp1[i+1][j+1], dp1[i][j]*A[i])
                    dp0[i+1][j+1] = max(dp0[i+1][j+1], dp1[i][j]*A[i])
    ans = dp0[N][M]
    print(ans)
