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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.Random;
import java.util.Scanner;

public class B {

	public static void main(String[] srgs) {

		Scanner nik = new Scanner(System.in);
		int t = nik.nextInt();
		StringBuilder st = new StringBuilder();
		while (t-- > 0) {
			int n = nik.nextInt();
			long[] a = new long[n];

			for (int i = 0; i < n; i++) {
				a[i] = nik.nextInt();
			}

			Arrays.sort(a);

			long last = a[n - 1] * a[n - 2] * a[n - 3] * a[n - 4] * a[n - 5];

			long th = a[n - 1] * a[n - 2] * a[n - 3] * a[0] * a[1];

			long o = a[n - 1] * a[0] * a[1] * a[2] * a[3];

			long res = Math.max(last, Math.max(th, o));
			System.out.println(res);

		}
		System.out.println(st);

	}

	private static long get(int[] a, int j) {
		long p = 1;
		for (int i = a.length - 1; i >= a.length - j; i--) {
			p = p * a[i];
		}
		return p;
	}

}