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 java.io.*;
import java.lang.reflect.Array;
import java.util.*;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
import java.util.stream.Stream;

public class Project1 {
    public static void main(String[] args) throws IOException {
        Scanner in = new Scanner(System.in);
        int T = in.nextInt();
        for (int t = 0; t < T; ++t) {
            int n = in.nextInt();
            Map<Integer, List<Integer>> list = Stream.generate(in::nextInt)
                    .limit(n).collect(Collectors.groupingBy(x -> (int) Math.signum(x)));
            if (!list.containsKey(1)) list.put(1, new ArrayList<>());
            if (!list.containsKey(-1)) list.put(-1, new ArrayList<>());
            list.get(1).sort((o1, o2) -> o2 - o1);
            list.get(-1).sort(Integer::compareTo);
            long max = Long.MIN_VALUE;
            for (int k : Arrays.asList(0, 2, 4)) {
                if (list.get(-1).size() < k || list.get(1).size() < 5 - k) continue;
                long res = list.get(-1).stream().limit(k)
                        .mapToLong(x -> (long) x)
                        .reduce((x, y) -> x * y).orElse(1) *
                        list.get(1).stream().limit(5 - k).mapToLong(x -> (long) x)
                                .reduce((x, y) -> x * y).orElse(1);
                if (res > max) max = res;
            }
            if (max == Long.MIN_VALUE && list.containsKey(0)) {
                System.out.println(0);
            } else if (max > 0) {
                System.out.println(max);
            } else {
                long res = -Stream.concat(list.get(-1).stream(), list.get(1).stream())
                        .map(x -> (long) Math.abs(x)).sorted()
                        .limit(5).reduce((aLong, aLong2) -> aLong * aLong2).orElse(Long.MAX_VALUE);
                System.out.println(res);
            }
        }
    }
}

